A game is played on an ${n \times  n}$  chessboard. At the beginning there are ${99}$  stones on each square. Two players ${A}$  and ${B}$  take turns, where in each turn the player chooses either a row or a column and removes one stone from each square in the chosen row or column. They are only allowed to choose a row or a column, if it has least one stone on each square. The first player who cannot move, looses the game. Player ${A}$  takes the first turn. Determine all n for which player ${A}$  has a winning strategy.
1. **Claim**: The game will end when all cells are empty.

2. **Proof by Contradiction**: Assume for the sake of contradiction that the cell with coordinates \((a, b)\) (in the \(a\)-th row and \(b\)-th column) is not empty but the game has ended. This implies that in row \(a\) and column \(b\), there must be at least one empty cell in each (otherwise, the game would continue). Assume \((a, d)\) and \((c, b)\) are these empty cells.

3. **Define Variables**: Let \(r_i\) and \(c_i\) denote the number of times row \(i\) and column \(i\) have been chosen during the game, respectively.

4. **Establish Inequalities**:
   - Since \((a, b)\) is not empty, \(r_a + c_b < 99\).
   - Since \((a, d)\) is empty, \(r_a + c_d = 99\).
   - Since \((c, b)\) is empty, \(r_c + c_b = 99\).

5. **Summing the Equations**:
   \[
   r_c + c_b + r_a + c_d = 2 \times 99
   \]
   This implies:
   \[
   r_c + c_d > 99
   \]
   This is a contradiction because \(r_c + c_d\) should be at most 99 (since each row and column can be chosen at most 99 times).

6. **Conclusion**: Therefore, the game must end when all cells are empty, and there will be exactly \(99n\) turns in total.

7. **Determine Winning Strategy**:
   - If \(n\) is odd, then \(99n\) is odd, and since Player \(A\) takes the first turn, Player \(A\) will also take the last turn (the 99n-th turn), thus winning the game.
   - If \(n\) is even, then \(99n\) is even, and Player \(B\) will take the last turn, thus winning the game.

The final answer is \( \boxed{ n } \) is odd.