Roughly equal number of swimmers in teams $b^2$ swimmers are to be put into one of the teams $1,2,\dots,b$. A team $i$ has a value function $f_i$, so that if they get swimmer $k$, they get value $f_i(k)$. The value $f_i(k)$ is randomized uniformly from $[0,1]$, independently of this value for other $i,k$. (So, there are $b^3$ different values in total.) The value that a team has for a set of swimmers is simply the sum of the individual values.
Suppose that we divide the swimmers to maximize the product of the team values. For large enough $b$, is it true that each team will get a number of swimmers in the range $[b/2, 2b]$ with high probability? In other words, is it highly likely that the swimmers will be divided roughly equally between the teams?
 A: This is not exactly a solution, but what seems a good way to approach the problem.
1) Consider for each player $k$ the team $I(k)$ to which he brings a maximal value 
$$
v(k)=\max_i f_i(k) = f_{I(k)}(k).
$$
BTW: note, that most probably for most of the players $v(k)$ is quite close to 1, as it is a maximum of $b\gg 1$ independent $R[0,1]$'s; to be more precise, the expectation of $v(k)$ is $1-\frac{1}{b+1}$.
2) Take a "greedy" way of forming the teams: put each player $k$ into the team $I(k)$ where he plays the best. Then, each player equiprobably goes to any of the teams, hence the number of players in any team is equal to the sum of $b^2$ expectation-$1/b$ Bernoulli variables, and thus is roughly normal with expectation and dispersion $\sim b$. In particular, the teams formed in this way are roughly equal, differing from $b$ by something like $\sqrt{b}$.
If I'm not mistaken, this means that the product for this case differs (in average, most probably) from the theoretical maximum of $b^b$ by a factor of constant. (This could be surprising, as in the theoretical upper bound 1 is used instead of all the values, but in fact the greedy method almost gets 1 everywhere, and $(1-1/b)^b$ is $1/e$ -- a constant)
3) The idea now is that if you divide the players in a strongly non-equal way, you will get a value that will be lower than the ``greedy'' one. 
4) Namely: imagine, that one of the teams is much smaller than the average, that it has less than $\epsilon b$ players (where, say, $\epsilon=\frac{1}{10}$). Then, you get an upper bound for the product by $\epsilon b$ times $(b+(1-\epsilon)b\cdot \frac{1}{b-1})^{b-1}$ (it is a bit rough: we're again setting all the players to values 1). But this is a theoretical maximum of $b^b$, multiplied by a constant, which is approximately equal to $\epsilon \cdot \exp(1-\epsilon)$, and the smaller is $\epsilon$, the smaller it becomes. In particular, for sufficiently small $\epsilon$ it becomes smaller, than the constant that we are getting in the greedy algorithm.
The same applies if one of the teams is too large, larger than $Ab$. The upper bound is again $b^b$ times $A\cdot \exp(1-A)$, and the factor $A\cdot \exp(1-A)$ tends to zero as $A$ tends to infinity.
5) Finally, it looks quite plausible that one can make the above arguments work for $\epsilon$ and $A$ arbitrarily close to $1$, but for that one should improve the above arguments in two points:
*) First, take $S=\sum_k v(k)$ to be maximal possible sum of values, and use $(S/b^2)^b$ as a reference point instead of $b^b$.
*) Second, re-equilibrate the teams: for every player, consider the team which is second-best for him, and try moving $\sim \sqrt{b}$ players from large teams to smaller ones, which are second-best for them.
It looks plausible that with this improvement of the greedy algorithm you get a product that is equivalent to our new maximum-reference point $(S/b^2)^b$, while any $\epsilon<1$ or $A>1$ reduce maximum possible value by a constant factor.
