The answer is all integers greater than or equal to $3$.
Obviously, it is impossible for $1$ and $2$.
If we can choose $n$ numbers from $1, 2$ and $-3$ such that they add up to $0$, then we fill an $n \times n$ table by filling the first row with those $n$ numbers, then each following row we "shift" one space to the right, i.e., the first term becomes the second, the second term becomes the third, and the last term becomes the first until the entire table is filled.
For $n \geq 3$, this is possible. For $3 | n$, just use $\{1,2,3\}$ $\frac{n}{3}$ times. For $n \equiv 1 \bmod{3}$, just use $\{1,1,1,-3\}$ once then use $\{1,2,-3\}$ $\frac{n - 4}{3}$ times. Lastly for $n \equiv 2 \bmod{3}$, just use $\{2,2,2,-3,-3\}$ once then use $\{1,2,-3\}$ $\frac{n-5}{3}$ times.