In each square of an $11\times 11$ board, we are to write one of the numbers $-1$, $0$, or $1$ in such a way that the sum of the numbers in each column is nonnegative and the sum of the numbers in each row is nonpositive. What is the smallest number of zeros that can be written on the board? Justify your answer.
1. **Define the problem constraints:**
   - We need to fill an $11 \times 11$ board with numbers $-1$, $0$, or $1$.
   - The sum of the numbers in each column must be nonnegative.
   - The sum of the numbers in each row must be nonpositive.

2. **Analyze the sum conditions:**
   - Let $C_i$ be the sum of the $i$-th column and $R_i$ be the sum of the $i$-th row.
   - We have the conditions:
     \[
     \sum_{i=1}^{11} R_i \leq 0 \quad \text{and} \quad \sum_{i=1}^{11} C_i \geq 0
     \]
   - Since the total sum of all entries in the board is the same whether summed by rows or by columns, we have:
     \[
     \sum_{i=1}^{11} R_i = \sum_{i=1}^{11} C_i
     \]
   - Combining these, we get:
     \[
     \sum_{i=1}^{11} R_i = \sum_{i=1}^{11} C_i = 0
     \]

3. **Implications of the sum conditions:**
   - Each row sum $R_i \leq 0$ and each column sum $C_i \geq 0$.
   - Since the total sum of all rows and columns is zero, each row sum $R_i$ must be exactly zero, and each column sum $C_i$ must also be exactly zero.

4. **Determine the structure of the board:**
   - For each row and each column to sum to zero, the number of $1$s must equal the number of $-1$s in each row and each column.
   - Since the sum of a row or a column is zero, the number of $1$s and $-1$s must be even.

5. **Minimum number of zeros:**
   - To satisfy the conditions, each row and each column must contain at least one zero. This is because if a row or column had no zeros, it would be impossible to balance the $1$s and $-1$s to sum to zero.
   - Therefore, there must be at least $11$ zeros on the board (one in each row and one in each column).

6. **Construct an example:**
   - Place a zero on the main diagonal of the $11 \times 11$ board.
   - Color the board like a chessboard, with alternating black and white squares.
   - Place $1$s on the white squares that do not contain a zero and $-1$s on the black squares that do not contain a zero.

By this construction, each row and each column will have a sum of zero, and there will be exactly $11$ zeros on the board.

The final answer is $\boxed{11}$.