Least number of zeros for an 11×11 $\{1,0,-1\}$ matrix with row/column sum conditions 
An $11\times11$ matrix can be only filled up with $1$s, $0$s and $-1$s. Find the least number of 0 you can write so that the sum of entries in every row is nonpositive and the sum of entries in every column is nonnegative.

My guess is that at least 21 zeros will be needed, but I'm having trouble proving it.
 A: Let the row sums of the matrix be $r_i$ and the column sums $c_i$, $i\in\{1,2,\dots,11\}$. We have
$$\sum_ir_i=\sum_ic_i$$
since both sides sum up all entries in the matrix. Since $r_i\le0$, $\sum_ir_i\le0$, and similarly $\sum_ic_i\ge0$. But this implies $\sum_ir_i=\sum_ic_i=0$ and hence $r_i=c_i=0$, i.e. all the row and column sums are zero.
For each column, if we use only $+1$ and $-1$ we cannot get the sum to be zero because of parity; we need at least one zero. Hence we need at least 11 zeros for the whole matrix, and this is indeed achievable as illustrated by the explicit construction below.
$$\begin{bmatrix}
0&+1&-1&+1&-1&+1&-1&+1&-1&+1&-1\\
-1&0&+1&-1&+1&-1&+1&-1&+1&-1&+1\\
+1&-1&0&+1&-1&+1&-1&+1&-1&+1&-1\\
-1&+1&-1&0&+1&-1&+1&-1&+1&-1&+1\\
+1&-1&+1&-1&0&+1&-1&+1&-1&+1&-1\\
-1&+1&-1&+1&-1&0&+1&-1&+1&-1&+1\\
+1&-1&+1&-1&+1&-1&0&+1&-1&+1&-1\\
-1&+1&-1&+1&-1&+1&-1&0&+1&-1&+1\\
+1&-1&+1&-1&+1&-1&+1&-1&0&+1&-1\\
-1&+1&-1&+1&-1&+1&-1&+1&-1&0&+1\\
+1&-1&+1&-1&+1&-1&+1&-1&+1&-1&0\\
\end{bmatrix}$$
Hence the least number of zeros required is 11.
