Answer: $32$

We first make a few observations.

* The sum of the numbers on all cards is $0$. Hence each row sum being nonnegative is equivalent to each row sum being $0$.
* For the same reason as above, each column sum must be $0$.
* The card $-4$ cannot be placed in a corner cell. Indeed, if $-4$ is placed in a corner cell as shown below, then the sum of the other two numbers in the first row is $4$, the sum of the other two numbers in the first column is $4$ and the sum of the other two numbers in the diagonal containing $-4$ is at least $4$. This is impossible since the sum of the positive numbers on the cards is only $1 + 2 + 3 + 4 = 10$.

* In essentially the same way we can prove that the card $-4$ cannot be placed in the central cell.

From the above observations, there are $4$ choices for the position of $-4$. After the position of $-4$ is fixed as shown below, there are $4$ choices for the other two numbers in the first row (we either put in $\{0, 4\}$ or $\{1, 3\}$ in either order), and then $2$ choices for the other two numbers in the second column (we put in the remaining pair in either order).

$4 \times 4 \times 2 = 32$

ways to fill in five cells. There is then only one way to fill in the remaining cells using the numbers $-3$, $-2$, $-1$ and $2$. For instance, suppose we have filled the cells in the following way, with $a$, $b$, $c$, $d$ denoting the numbers yet to be filled in:

<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>a</td><td>0</td><td>b</td></tr><tr><td>c</td><td>4</td><td>d</td></tr></table>

By considering the rows, we need $a + b = 0$ and $c + d = -4$, so $\{a, b\} = \{2, -2\}$ and $\{c, d\} = \{-1, -3\}$. By considering the columns, we need $a + c = -1$ and $b + d = -3$. Hence the only possibility is $(a, b, c, d) = (2, -2, -3, -1)$. It remains to check whether these $32$ ways also satisfy the requirement that the two diagonal sums are nonnegative. We start with the following $4$ cases with the position of $-4$ fixed:

<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>2</td><td>0</td><td>-2</td></tr><tr><td>-3</td><td>4</td><td>-1</td></tr></table>

<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>-3</td><td>4</td><td>-1</td></tr><tr><td>2</td><td>0</td><td>-2</td></tr></table>

<table><tr><td>0</td><td>-4</td><td>4</td></tr><tr><td>2</td><td>1</td><td>-3</td></tr><tr><td>-2</td><td>3</td><td>-1</td></tr></table>

<table><tr><td>0</td><td>-4</td><td>4</td></tr><tr><td>-2</td><td>3</td><td>-1</td></tr><tr><td>2</td><td>1</td><td>-3</td></tr></table>

All cases have both diagonal sums being nonnegative. The other $4$ cases with $-4$ in the same position must therefore also have the same property, since they are obtained by swapping the first and third columns from the $4$ cases above:

<table><tr><td>3</td><td>-4</td><td>1</td></tr><tr><td>-2</td><td>0</td><td>2</td></tr><tr><td>-1</td><td>4</td><td>-3</td></tr></table>

<table><tr><td>3</td><td>-4</td><td>1</td></tr><tr><td>-1</td><td>4</td><td>-3</td></tr><tr><td>-2</td><td>0</td><td>2</td></tr></table>

<table><tr><td>4</td><td>-4</td><td>0</td></tr><tr><td>-3</td><td>1</td><td>2</td></tr><tr><td>-1</td><td>3</td><td>-2</td></tr></table>

<table><tr><td>4</td><td>-4</td><td>0</td></tr><tr><td>-1</td><td>3</td><td>-2</td></tr><tr><td>-3</td><td>1</td><td>2</td></tr></table>

The other cases (where the position of $-4$ is different) can be obtained by rotating one of the above configurations. Hence all $32$ ways of filling in the table in which all row and column sums are zero will also have both diagonal sums being nonnegative. It follows that the answer is $32$.