Edit 2:
The 189-plane solution assumed that planes could return to base, refuel, and take off again. If this is not allowed, the math gets simpler, to the point that we can get an almost-analytic solution. Let $D(n_0)=\sum_{i=1}^{n_0}\frac{1}{i}$ be the distance a fleet of $n_0$ planes can travel before its last plane runs out of fuel. Also let $R(n_1,n_2,...)$ be the distance from which a plane can be retrieved by a sequence of rendezvous fleets of sizes $n_1, n_2, ...$ To make it around a globe of size 10, we need $D(n_0)+R(n_1,n_2,...)\geq 10$, and the goal of the problem is to find the minimum $N=\sum_i n_i$ that makes that possible.
To retrieve a plane at distance $R$, the $n_1$ fleet must fly out the full distance $R$, and then enough distance back to get them within $R(n_2,n_3,...)$ before running out of fuel. As a recursion relation, $$R(n_1,n_2,...)=D(n_1)-R(n_1,n_2,...)+R(n_2,n_3,...)$$$$\rightarrow R(n_1,n_2,...)=\frac{1}{2}\left(D(n_1)+R(n_2,n_3,...)\right)$$$$\rightarrow R(n_1,n_2,...)=\sum_{i=1}^{imax} \left(\frac{1}{2}\right)^iD(n_i)$$$$\rightarrow D(n_0)+R(n_1,n_2,...)=\sum_{i=0}^{imax} \left(\frac{1}{2}\right)^iD(n_i)$$
If $n_i$ could be continuous, we would maximize this for a given $N$ by equalizing the derivatives with respect to all $n_i$; this results in $n_i=N/2^{i+1}$. Since $n_i$ is discrete, the optimum takes a little more fiddling to find, but we still expect $n_i\approx n_{i-1}/2$. Whether by exactly solving the discrete difference equation or with a little numerical searching, we can get a solution with 328 planes, split into fleet sizes 166, 83, 41, 20, 10, 5, 2, 1.
Edit:
Incorporating the tips in the comments, we can do a lot better yet. Compared to my original answer below, we can increase efficiency with two insights:
Even if a given plane isn't going to fly out with a fleet and go until its tank is empty, it is a waste to leave it parked in the base. Instead, it should fly out briefly with the fleet, supply fuel for as long as it can, and then fly home; this gives the fleet a little extra fuel at no extra cost. More precisely, if we have $N_{\textrm{tot}}$ planes and we want $N_1$ of them to fly 'til they drop, the first plane can use $\frac{N_{\textrm{tot}}}{N_{\textrm{tot}}+1}$ fuel to supply the fleet for a distance $\frac{1}{N_{\textrm{tot}}+1}$ before turning back and using its last fuel to get home. The second plane can then use $\frac{N_{\textrm{tot}}-1}{N_{\textrm{tot}}+1}$ to supply the fleet for the same distance, and use $\frac{2}{N_{\textrm{tot}}+1}$ to get all the way home. Conveniently, every plane that flies partway out and returns in this way gives us the same amount of extra distance; all together, the planes that would otherwise stay home can carry the fleet out to $d=\frac{N_{\textrm{tot}}-N_1}{N_{\textrm{tot}}+1}$. This not only directly increases the distance covered for a given $N_1$ and $N_2$, but increases the optimal $N_2$ for a given $N_{\textrm{tot}}$.
We don't have to stick with just one rendezvous fleet. When the rendezvous planes set off to meet the main fleet, some fraction of them can go part of the way, peel off, and fall back to refuel following the above strategy.
Using both these improvements, and taking the time to do a more careful numerical search, I can find a solution with 189 planes. All 189 planes launch together, and 117 of them carry the fleet to $d=0.615$ before returning to base. The remaining 72 planes fly, crashing as they run out of fuel, to $d=5.476$. The 117 planes that returned to base then escort the last plane from the original fleet home, over the course of 10 ever-smaller rendezvous flights; the sequence of fleet sizes at takeoff is [117, 72, 44, 27, 16, 10, 6, 4, 2, 1].
Original Answer:
Following the previous answer in this thread, let's say that each plane has a fuel capacity of 1, which is enough for it to travel a distance of 1, meaning the trip around the world covers a distance of 10. Since each plane has the ability to share its fuel with any number of other planes mid-flight, we can keep track of the fuel capacity and consumption of an entire fleet of planes, without worrying about how much fuel any individual plane has in its tank.
A fleet of $N$ planes has a total fuel capacity of $N$, and uses $N$ fuel per unit distance traveled. To minimize fuel consumption, the size of the fleet should always be big enough to hold all the available fuel, but no bigger. One way to accomplish this is to have, at any given time, a single plane using its fuel-sharing magic to power the entire fleet; once that plane is out of fuel, it crash-lands and another takes its place. Using this strategy, an $N$-plane fully-fueled fleet (FFF) can cover a distance of $\frac{1}{N}$ before one plane crashes and it becomes an $(N-1)$-plane FFF. Generalizing to multiple steps, the minimum fleet size $N(d)$ that can cover a distance $d$ is
$$N(d)=\textrm{minimum }N\textrm{ such that }\sum_{i=1}^N\frac{1}{i}\geq d$$
If all the planes had to circle the globe one-way, our best solution would be $N(10)=12367$. But we can do better by starting off with one fleet big enough to get most of the way around the world, then sending a second fleet to meet it from the other direction. If the fleets meet up at some distance $d_m$, we have two requirements: the initial fleet size must be big enough to cover $d_m$, and the second fleet size must be big enough to cover the remaining distance $10-d_m$ as a round trip. Summing these to get the total number of planes:
$$N_{\textrm{total}}=N(d_m) + N\big(2(10-d_m)\big)$$
I don't know of a good way to solve these inequalities analytically, but with some numerical trial and error, I found a solution with 840 planes total. We send out 514 planes, which fly until one plane with a nearly empty fuel tank reaches the rendezvous point at $d\approx6.82$.Another fleet comes from the other direction with 326 planes, 13 of which reach the rendezvous point. The remaining 13 planes can barely make it the rest of the way to $d=10$.
