Update:
I made a big blunder in the previous version, this should be fixed now.
Part 1: It is always possible to remove amount $x$ from any number $n \ge 7$ of pots if all pots have at least amount $x$
Let $p$ be the number of permutations with 7 pots containing a specific pot. Cycle all permutations and remove amount $\frac xp$ from each pot. As each pot occurs $p$ times the total removed amount from each pot is $x$.
Part 2: A combination of pots where the 7 pots containing the most amounts are equal is know to work regardless of the amounts in other pots
Consider the following image:
The bars show the amounts in the pots, ordered by the amount. The amounts in the first 7 pots are equal. Start with the smallest pot $J$ and remove the amount of $J$ from all pots (this works as seen in part 1). Repeat with pot $I$ and then with pot $H$. At the end we have the top 7 pots containing only the black marked area, which can be emptied in the next step.
Part 3: Removing superfluous amount in biggest pot requires 6 times the same amount 20 smallest pots
Consider the following image:
We must remove the red part in pot $A$ to reach the situation from part 2. It is easy to see that we can do this easily if we have the same 6 parts in pots $H-M$.
This is the place where I made a blunder in the previous version, assuming I could trade the red part in $A$ with a combination of parts of the same size in pots $>G$, instead of 6 times the size.
What can be said about the distribution of the amounts beyond pot $G$? Actually it doesn't matter, if we use a trick. Consider the following image.
We want to remove the red part in pot $A$. The orange part in pot $H$ has 6 times the size of the red part. We can remove the red part and the orange part of pot $H$ while also removing the orange parts in pots $A-G$. To do this we generate all permutations of 6 pots from the range $B-H$:
$\binom{7}{6}$ = 7
The number each of the pots $B-H$ appears in this permutations is:
$\binom{6}{5}$ = 6
Now we add pot $A$ to all these permutations and remove the red part from each permutation. We remove the red part 7 times from pot $A$ and 6 times from pots $B-H$ which is exactly what we see in the image above. This can be easily extended to any number of pots, as long as they contain 6 times the size of the red part.
The other conclusion is, no matter what we do, we can't remove the red part if the size of the pots beyond $G$ is less than 6 times the size of the red part.
Part 4: Removing superfluous amount in 2 biggest pot requires 2.5 times the same amount 20 smallest pots
Consider the following image:
Similar to part 3, we now want to remove the 2 red parts in pots $A-B$. It is again easy to see that we can do this easily if we have the same 5 parts in pots $H-L$.
Using the same reasoning as in part 3, it can be shown, that the amount distribution beyond pot $G$ doesn't matter.
It can also be shown that similar applies to superfluous amounts in pots $C-F$.
Part 5: The general case
The following image shows the general case which is guaranteed to be solvable based on the parts above:
As we have seen in part 3, the distribution in pots beyond $G$ doesn't matter. Also as seen in part 2 adding amounts in pots beyond $G$ is solvable too. What is left is to show which amounts can be solved in any case.
Part 6: The edge case
The above image shows the edge case, where the red area is maximized (if we limit ourselves to 13 pots). The height of pots $H-M$ must be the same as the height of the red part of pot $A$. Pots $B-G$ must have also the same size. As $A$ is limited to $1$, the height of pots $B-M$ must be $\frac{1}{2}$. The size of this case is:
$1 + 12 * \frac{1}{2} = 7$
This means that 7kg honey is enough to guarantee, that the puzzle is solvable, if the pot size is limited to 1kg.
Any other amounts which we would try to add, don't change anything. Amounts added to pots beyond $G$ are OK as shown in part 2. Amounts added to pots $B-G$ reduce the required amount in pots beyond $G$ as can be seen in the following image:
One more thing we could do, is to use the other 14 pots we didn't use yet to further maximize the red area, like in the following image:
We used here one more pot. We increased the size of the red area above $A$ by some amount $x$. We also increased the size of the red area beyond pot $G$ by $6x$. But at the sime time we decreased the size of the black area by $7x$ which in the end gives a total change of $0$.
The pot size which would still work for 17kg is (based on the first edge case):
$x + 12 * \frac{x}{2} = 17$
$x = \frac{17}{7} = 4,28...$
Also, this solutions works with any number of pots $\ge 13$.
Alternatively we can make Winnie-the-Pooh really happy and tell him, he can eat from 17 pots each day (which would still work with 17kg and 1kg max per pot).