A table with $m$ rows and $n$ columns is given. At any move one chooses some empty cells such that any two of them lie in different rows and columns, puts a white piece in any of those cells and then puts a black piece in the cells whose rows and columns contain white pieces. The game is over if it is not possible to make a move. Find the maximum possible number of white pieces that can be put on the table.
1. **Claim**: The maximum number of white pieces that can be placed on the table is \( m + n - 1 \).

2. **Construction**: 
   - Start by placing a white piece in the top-left corner.
   - Continue placing white pieces in cells that are either in the top row or the leftmost column.
   - This ensures that each white piece is placed in a unique row and column, and no two white pieces share the same row or column.

3. **Induction Base Case**:
   - If \( m = 1 \) or \( n = 1 \), the result is clearly true. For example, if \( m = 1 \), we can place \( n \) white pieces in the single row, and similarly for \( n = 1 \).

4. **Induction Hypothesis**:
   - Assume the result holds for all tables with dimensions \( m' \times n' \) where \( m' + n' < m + n \).

5. **Induction Step**:
   - Consider the set of white pieces at the end of the game. No four white pieces can form a rectangle because it would imply that we could place white pieces on all four vertices of the rectangle, which is not possible.
   - Define a white piece as a **row loner** if it is the only white piece in its row, and a **column loner** if it is the only white piece in its column.
   - If there exists a row loner or a column loner, remove that row or column with only one white piece and apply the induction hypothesis to the remaining \( (m-1) \times n \) or \( m \times (n-1) \) table.

6. **Contradiction**:
   - Suppose no white piece is alone in its row or column. Consider the last white piece placed.
   - None of the white pieces in its same row or column can be placed during the same move.
   - Since there is at least one other white piece in its row and in its column placed before it, the cell where this last white piece was placed would have been black, leading to a contradiction.

Thus, the maximum number of white pieces that can be placed on the table is \( m + n - 1 \).

The final answer is \( \boxed{ m + n - 1 } \).