Answer:$11$. Let $C_i$ be $i.$ column and $R_i$ be $i.$ row. We have $\sum R_i \leq 0 \leq \sum C_i=\sum R_i$ which means $\sum R_i=\sum C_i=0$ $\implies C_i=0=R_i$ The sum of a row or a column is even. There should be atleast one $0$ at all columns and all rows. So there are atleast $11$ times $0$. Example: One of the main diagonal is full of $0$. Then we color the board as a chess board. We put $1$ at white squares which does not contain $0$ and $-1$ at black squares which does not contain $0$.