Well, I spent a lot of time to make this out. But now I have got the solution ready. It's not spoilerware, because it's too long (moderation, anyone?)
220 transmissions to win, 230 - to inform the Agent.
But first I will restate a bit simpler version of 270(269?) transmissions' solution, to get those of you, who didn't get it, into the parcel business.
I use term "delivery score" (ds) as amount of parcels a given agent gets. I also use "maximum delivery score" as a maximum ds among all agents (thus, starting at mds = 0). Due to Post Machine's nature, every ds at any given moment equals either mds, or mds-1. I also use 00X as "any agent except for 001", but that's obvious.
Now let us look at the first modelled situation: agent 00X got a box, and he wants to give it to agent 00Y. All the other agents know, that the package is intended for 00Y. 00X just throws a package in the mail, and all other agents resend it as soon as they get it. Then the pricnciple is: if mds=N at the beginning, then when 00Y gets the box, mds will be no higher then N+1 (guaranteed).
This fact is rather obvious. First, all the agents will get ds=N. Then, the box will travel among all of them, before ds becomes N+1 for all -- thus, it will reach 00Y in time.
Retrace of 270-solution
Original "270"-solution is overcomplicated with varying order of agents. Let's make it even simpler: what if we wanted the data (interchanging the boxes) travel in a predefined order: 001-002-001-002-003-002-003-004-003-004...(please refer to original solution for locking operations). Well, we CAN do that for the same price. It's 27 travels, thus making mds no more than 27, giving us 270 steps.
Now why during "270"-solutions other agents won't take boxes for theirs? The author suggests playing with boxes' sizes. I think I have an easier solution: any agent can hang a corresponding amount of locks on a box (like, 1, 2, etc.) to mark to whom he is sending the box. Thus, agent 007 finding box with six padlocks will pass it further, and finding a box with 7 padlocks - will hang 10 more to make sure nobody else locks it and send back to 006.
So this is the 270-solution (I still don't get 269 part - I think it's a mistake or different interpretation of the problem). Remember two things from here - marking boxes with special amount of locks and common "resending a single box mechanics" (also, mds rule). It will come in handy, but not too soon.
I ALSO HAVE TO MENTION that, though 270 transmissions DO solve the problem, it is Agent 010, who will be informed that the problem has been solved. Agent 1 may still false-report that "problem's been solved" while the packages are still on the way (or vice versa). To complete the solution, agent 010 has to hand out a signal: 10 packages with "finished" written in locks, or, better, one package with "please resend this to 001 i am in good health and almost out of padlocks" written with locks). Thus, this solution is actually 280 transmissions.
That. Was. The warm-up. We're getting to the serious spy tricks. The following solution will take 220 transmissions for mailing system to be sure of mission completion and 230 transmissions to moreover inform the Agent 001 of mission completion.
Phase (1) - Distribution.
Agent 001 gathers 10 packages with copies spy-data, locks each with a lock and sends them by mail. It is definite that every agent (including 001) will get a package. Agents 00X should hang their private padlocks on those packages as soon as they get 'em. Agent 001 opens the package he send to himself, reads the spy-data again and saves the padlock for a better use. Now the ds for every agent is 1, or mds=exact 1.
Also, every agent already got a double-locked parcel. Fast work!
Phase (2) - Chaotic retrieval.
Agent 001 gathers a blackbox (from here on "blackbox" is a box full of ink with exactly 666 locks on it. It is a common spy expression). Agent 001 sends a blackbox.
Now we have a little bit of ambiguity, as blackbox may be received by agent 00X or agent 001 himself. Let's look at the strategy of agents 00X for a minute.
Agent 00X is instructed to hold on to his first delivery and put a lock on it (see phase 1). He is also instructed: as soon as you get a second parcel please desperately try to send all parcels with two locks away.
Thus, as soon, as an agent gets a blackbox, he sends his original parcel away. Whenever he hits another agent 00Y with a two-locked parcel, he also resends these two. And etc., without any order.
But - you may think - what if agent 001 sends blackbox to 002, and 002 sends his parcel back to 001, without making contact with other agents, thus, without making them fall into chaos?
Well, exactly for that case, agent 001 will not only send blackbox, but send away his second delivery, and after that accept all the deliveries that come to him. As you can check this is enough to get "second parcel panic" signal to all of the agents (if agent 001 already got second delivery, others are guaranteed to get it before anything else gets sent to 001).
Outcome of phase 2 :
Agent 001 has all the 9 agents' packages. We know they won't stop resending them until they get rid of them completely. Thus, the chaos will stop only after agent 001 gets all the goods.
Agent 001 has exactly 11 delivery score. 1 is got during phase 1. 2 is the package he resent to get the gears running (also you might notice - delivery 2 could be his blackbox delivered back to him by the mail). Deliveries 3 to 11 are 9 packages he got from agents during the chaotic phase.
Every other agent has 10 to 11 delivery score (read as: mds = 11).
Phase (3) - Housework.
Agent 001 has to find 9 keys and fit them to 9 locks. It's the most tedious phase, but it takes no transmissions.
Phase (4) - Rebalancing.
-- Oh 'e gods! - says agent 001 - I want to conduct distribution step again. But I realised, that some (maybe none) of the agents may have 10 delivery score and some - 11. I can't just send these nine boxes, for two may reach the same person! My plan is ruined! But what if... I were to make delivery score the same for every agent? Ho-ho-ho, delightlfully devilish, agent 001.
All other agents, who now got rid of the boxes, wait patiently for agent 001's move. Agent 001 creates a whitebox (it is the one with 1000 locks on it and filled with milk) and sends it throught the net. Now other agents act according to plan:
If agent 00X with 10 delivery score gets the whitebox, he resends it.
If agent 00X with 11 delivery score gets the whitebox (possibly: for the second time), he slams a new lock onto it and then resends it.
When agent 001 receives the whitebox back, he counts the locks carefully. Suppose it is 1000+k. Then all the agents have at least 11 score and there is exactly k agents (besides 001) with exactly 12 score (they got +1 for box at eleven score, and each added a lock). Now...
Now agent 001 throws whitebox away (not to confuse agents), gets his supply of black boxes and sends (9-k) boxes. He is sure that they will get to remaining 11-ds agents. He is also sure, that the agents are instructed not to react to a blackbox of 12th delivery.
-- Phew! -- says agent 001. -- Now every agent (including myself) has a score of 12. I hope you're ready for more confusing steps.
Phase (5) - Distribution.
Agent 001 still got the boxes, each with a single lock. He adds a blackbox to them and sends all 10. Then he waits for the return. Every agent's ds is 13 now. 9 packages are distributed evenly - one for each, and one without a package. Maybe 001, maybe not.
Now we get back to the original method of solution - consequential single-box transmission.
Phase (6) - Order.
Let me assume that 001 got the blackbox, and then get to the case where he got a usual, one-locked box back.
Agent 001 packs the blackbox and sends it into the net. The following algorithm is pretty clear, and you might understand that it works, but the fact is: it works fast. Now any agent (including 001) reacts to a box as follows:
If it is a blackbox and 00X has a locked box on hands - 00X takes the blackbox and sends the locked box.
If it is a blackbox and 00X has got nothing - 00X just resends the blackbox.
If it is a usual box (with one key), 00X tries to open it. In case of failure -- he resends it. In case of success -- he either sends another usual box (if he's got one) or a blackbox.
I tell you, that agents will all get the data in +9 mds. They wouldn't know it though, but that's another question.
Why is so?
Well. Any permutation (and we're dealing with permutation here) is a combination of cycles. I'm telling that, if the traveling box is a blackbox, then for every cycle either every person in it knows the data, either every person doesn't. It's called induction - check it for starting position (nobody, except for lucky one-cyclers, know the data) and continue to check further.
Imagine the box is black and has been just sent. Mds is N (13, for instanse). Who will replace it with a usual box? A person from 'unsatisfied cycle' will (call him Carl). Moreover, Carl will do it before other people of the cycle get the box and ds-increase - because, if they had a box, they would also replace it.
And who needs Carl's box? Well, someone from his cycle - the one that didn't get the blackbox. It means that he (...Bert?) will also increase ds by one, but only when he gets Carl's box. Two operations - sending a blackbox, swapping it, and delivering Carl's box to Bert - take +1 mds alltogether. It's... hard to comprehend, at least for me.
What happens after Bert gets the box is easy - he replaces it, and it needs to be delivered to next person from the cycle. And we already know - it takes +1 mds at most (and nobody will open the box, because it's locked with a unique key). And so on, until Carl gets the box back and whole cycle is satisfied.
- If agent 001 got a usual box returned, he just sends the usual box. You can see for yourself, that the strategy will work just as fast.
Phase (7) Conclusion
Almost done. No, actually - done. mds = 13+9 is 220, we know for sure, that after 220 transmissions we'll get the data to everyone. Almost seems like a cheap trick to me. But why stop there?
221 transmissions are enough to stop. The one, who gets ds=22 may realise, that all is done and stop resending black box for eternity.
230 transmissions are enough to inform the 001 of success -- same stuff, but 001 stops resending after his ds is 22.
Also I'm 99% sure there are more tricks to make better solutions. Just looking "outside the box" (pun intended), I managed to find three completely perverted ideas to improve the answer. There must be more.