The answer is $2^{n-1}+1$. $A$ can inductively ensure that it's possible to get $2^{n-l}$ boxes which contain at least $l$ ballots for $l\in \{ 1,2...,n\}$. Now we need to show that if $m\leq 2^{n-1}$, $B$ can ensure that there won't be any box containing $n$ ballots. Note that when there's any box with at least $2$ ballots occurred after $B$ turn, after that, any boxes with no ballot can be consider "dead" (i.e. it'll be a waste turn for $A$ to choose them). So, the best that $A$ can do before any "dead" boxes occurred is to make $m-1$ boxes, each with one ballot. There is a box which is now dead, we won't consider it. Suppose that the number of ballots in $m-1$ boxes are $b_1,b_2,...,b_{m-1}$, let $S=\sum_{i =1}^{m-1}{2^{b_i}}$. WLOG $b_1\geq b_2\geq ...\geq b_{m-1}$. If $A$ chooses two boxes with $b_i,b_j$ ballots, we've two cases. First, if $b_i,b_j<b_1$, then if $B$ chooses the box with maximum number of ballots, which contain $b_1$ ballots, the new set of $m$ boxes will give $S'=S+2^{b_i}+2^{b_j}-2^{b_1}+2^0\leq S+1$. Second, if at least one of $b_i$ and $b_j$ is equal to $b_1$, WLOG $b_i=b_1$, then, again, $B$ chooses the box with maximum number of ballots, this time that box contain $b_1+1$ ballots, the new set of $m$ boxes will give $S'=S+2^{b_1}+2^{b_j}-2^{b_1+1}+2^0=S+2^{b_j}-2^{b_1}+1\leq S+1$. These two cases gives us $S$ will increase by one for each turn. But note that after each turn of $B$, there will be one more "dead" box. So there can be at most $m-2$ turns before $A$ can't chooses $2$ non-dead boxes. Note that the initial value of $S$ is $(m-1)\times 2^1=2(m-1)$ It's impossible to have a box contain at least $n$ ballots since $2^n+(m-2)\times 2^0=2^n+(m-2)>2(m-1)+(m-2)$.

The answer is $m = 2^{n - 1} + 1$. We first show that $m \geq 2^{n - 1} + 1$ must hold. To any position where the non-empty boxes have $a_1, a_2, \dots, a_r$ ballots, assign the sum $$S = \sum_{i = 1}^{r} 2^{a_i - 1}$$ We show that $B$ can play in such a way that $S$ does not increase, unless $A$ chooses two empty boxes and no box contains two or more ballots. If $A$ chooses two non-empty boxes with $r$ and $s$ ballots, assume $r \geq s$. After $A$'s turn, $S$ increases by $2^{r - 1} + 2^{s - 1}$, and $B$ may remove the box now containing $r + 1$ ballots, decreasing the value of $S$ by $2^r \geq 2^{r - 1} + 2^{s - 1}$. If $A$ chooses an empty box and a non-empty box with $r$ ballots, then by removing this box, which now contains $r + 1$ ballots, $B$ can make the value of $S$ decrease by $2^{r - 1} - 1 \geq 0$. Finally, if $A$ chooses two empty boxes, then $S$ increases by $2$. If some box contains more than one ballot then $B$ may remove it, decreasing the value of $S$ by at least $2$. Otherwise, $B$ removes a box with one ballot and $S$ increases by one. Finally, since $B$ always empties one box, $A$ may guarantee at most $m - 1$ boxes with one ballot. After this point $S$ may not increase and thus $S$ is bounded above by $m - 1$. If some box eventually contains $n$ ballots after some turn of $B$, then $S \geq 2^{n - 1}$, which combined with $S \leq m - 1$ gives $m \geq 2^{n - 1} + 1$. Now, if $m = 2^{n - 1} + 1$, then $A$ may guarantee a win by first repeatedly choosing two empty boxes until $2^{n - 1}$ of them contain exactly one vote. Then $A$ chooses pairs of boxes containing one ballot. $B$ may remove at most one box at a time and hence $A$ can obtain $2^{n - 2}$ boxes with $2$ ballots. Inductively, $A$ may obtain $2^{n - k}$ boxes with $k$ ballots, giving a box with $n$ ballots for $n = k$ as desired.

We show that $A$ cannot force a win for $m = 2^{n - 1}$ if $B$ always empties the box with the greatest number of ballots. We prove by induction on $k$ that after each of $B$'s turns there are at most $2^{n - k} - 1$ boxes with $k$ ballots. This is evidently true for $n = 1$ as $B$ empties a box after each of his turns. Now, we prove that for each $k \geq 2$, if $a$ boxes contain $k$ or more ballots then at most $2^{n - k + 1} - 2a - 1$ boxes contain exactly $k$ ballots. In any moment where $a = 0$ this is true because of the inductive hypothesis applied to $k - 1$. Now, assume by contradiction that this inequality becomes false at some moment, and consider the first moment where there are $a$ boxes with $k$ or more ballots, and at least $2^{n - k + 1} - 2a$ with exactly $k - 1$. Assume that, in her previous turn, $A$ chose $\ell$ boxes with exactly $k - 1$ ballots, and $2 - \ell$ with a different number of ballots. Then after her turn, the number of boxes with $k$ or more ballots increases by $\ell$, while the number of boxes with exactly $k - 1$ ballots increases by at most $(2 - \ell) - \ell = 2 - 2\ell$ (because at most $2 - \ell$ boxes with $k - 2$ ballots were chosen, and $\ell$ boxes with $k - 1$ ballots switched to having a different number of ballots). $B$ responds by removing a box with $k$ or more ballots and thus after both turns the number of boxes with $k$ or more ballots increases by $\ell - 1$. It follows that before these two turns, the number of boxes with $k$ or more ballots was $a - \ell + 1$, while the number of boxes with exactly $k - 1$ ballots was at least $2^{n - k + 1} - 2a - (2 - 2\ell) = 2^{n - k + 1} - 2(a - \ell + 1)$. This contradicts our assumption that the moment previously established was the first in which our inequality did not hold. This proves our inequality, and in particular this implies that $2^{n + k - 1} - 2a - 1$ is non-negative, and hence $a \leq 2^{n - k} - 1$ which completes the induction, and proves by plugging $k = n$ that $A$ cannot guarantee obtaining a box with $n$ ballots. The winning strategy when $m = 2^{n - 1} + 1$ is the same as in the previous solution.