Solution:
We put $m+n-1$ white pieces by $m+n-1$ moves consecutively in the cells $(1,1), (2,1), \ldots, (m,1), (1,2), (1,3), \ldots, (1,n)$.

We shall call a closed chain of cells such that at least two numbers of rows and columns change alternatively a zig-zag cycle.

Note that there is at least one black piece in the cells of a zig-zag cycle. Indeed, assume the contrary and consider the last white piece. Its neighbors (in the zig-zag cycle) are white pieces that have been put before it. Then the last piece must be black, a contradiction.

Assume now that the game is over and there are more than $m+n-1$ pieces. Remove a row or a column with at most one white piece, repeat the same operation, etc. Since such a removing can be done at most $m+n-1$ times, any row and column in the new table will contain at least two white pieces. This new table, and hence the given table, has a zig-zag cycle, a contradiction to the proved above.