Let $n \geq 2$ and $m$ be positive integers. $m$ ballot boxes are placed in a line. Two players $A$ and $B$ play by turns, beginning with $A$, in the following manner. Each turn, $A$ chooses two boxes and places a ballot in each of them. Afterwards, $B$ chooses one of the boxes, and removes every ballot from it. $A$ wins if after some turn of $B$, there exists a box containing $n$ ballots. For each $n$, find the minimum value of $m$ such that $A$ can guarantee a win independently of how $B$ plays.
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 \( \boxed{ m = 2^{n-1} + 1 } \)