1. **Establishing the minimum value of $m$ such that $A$ can guarantee a win:**

We need to show that $m \geq 2^{n-1} + 1$ must hold. To do this, we will use a sum $S$ defined as follows:
$$S = \sum_{i=1}^{r} 2^{a_i - 1}$$
where $a_i$ represents the number of ballots in the $i$-th non-empty box.

2. **Analyzing the changes in $S$ based on $A$'s and $B$'s moves:**

- If $A$ chooses two non-empty boxes with $r$ and $s$ ballots (assume $r \geq s$), then after $A$'s turn, $S$ increases by $2^{r-1} + 2^{s-1}$. $B$ can then remove the box with $r+1$ ballots, decreasing $S$ by $2^r$. Since $2^r \geq 2^{r-1} + 2^{s-1}$, $S$ does not increase.

- If $A$ chooses an empty box and a non-empty box with $r$ ballots, then $B$ can remove the box with $r+1$ ballots, decreasing $S$ by $2^r - 2^{r-1} = 2^{r-1}$. Since $2^{r-1} \geq 1$, $S$ does not increase.

- If $A$ chooses two empty boxes, $S$ increases by 2. If some box contains more than one ballot, $B$ can remove it, decreasing $S$ by at least 2. Otherwise, $B$ removes a box with one ballot, and $S$ increases by 1.

3. **Bounding $S$:**

Since $B$ always empties one box, $A$ can 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, then $S \geq 2^{n-1}$. Combining this with $S \leq m-1$, we get:
$$m-1 \geq 2^{n-1} \implies m \geq 2^{n-1} + 1$$

4. **Constructing a winning strategy for $A$ when $m = 2^{n-1} + 1$:**

- $A$ can first repeatedly choose two empty boxes until $2^{n-1}$ of them contain exactly one ballot.
- Then, $A$ chooses pairs of boxes containing one ballot. $B$ may remove at most one box at a time, so $A$ can obtain $2^{n-2}$ boxes with 2 ballots.
- Inductively, $A$ can obtain $2^{n-k}$ boxes with $k$ ballots, eventually leading to a box with $n$ ballots.

5. **Proving that $A$ cannot force a win for $m = 2^{n-1}$:**

- Assume $B$ always empties the box with the greatest number of ballots.
- By induction on $k$, we show that after each of $B$'s turns, there are at most $2^{n-k} - 1$ boxes with $k$ ballots.
- For $k \geq 2$, if $a$ boxes contain $k$ or more ballots, then at most $2^{n-k+1} - 2a - 1$ boxes contain exactly $k$ ballots.
- If this inequality becomes false, consider the first moment where there are $a$ boxes with $k$ or more ballots and at least $2^{n-k+1} - 2a$ with exactly $k-1$ ballots.
- After $A$'s turn, the number of boxes with $k$ or more ballots increases by $\ell$, and $B$ removes one, so the number of boxes with $k$ or more ballots increases by $\ell - 1$.
- This leads to a contradiction, proving that $A$ cannot guarantee a win for $m = 2^{n-1}$.

$\blacksquare$

The final answer is $\boxed{ m = 2^{n-1} + 1 }$