We will prove that the desired numbers $k$ are those that are not factors of $n$.

If $k \mid n$, then we can tile the $n \times n$ square with $k \times k$ squares and the total sum should simultaneously be positive and negative, which is impossible.

If $k \nmid n$, then $n = kq + r$, where $0 < r < k$. We fill the unit squares with $a$ (to be chosen conveniently later on) in the positions $(ik, jk)$ with $i, j = 1, \dots, q$ and with $1$ in the other positions. Every $k \times k$ square contains exactly one unit square of the form $(ik, jk)$, therefore the sum in every $k \times k$ square is $a + k^2 - 1$. The total sum is $q^2 a + n^2 - q^2$. We will choose $a$ arbitrarily from the non-empty interval $\left(1 - \frac{n^2}{q^2}, 1 - k^2\right)$.

Remark: For the case $k \nmid n$ there are many other ways of choosing the numbers from the unit squares. Another choice is to fill all the unit squares of the columns $jk$, $j = 1, \dots, q$, with a convenient $a$ and the other ones with $1$.