Suppose that an $n \times n$ table is filled with numbers $-1$, $0$, and $1$ in such a way that the conditions are met. The sums $n$ and $-n$ can be obtained only from a row, column or diagonal with all $1$s and all $-1$s, respectively, whence $n$ and $-n$ cannot arise as sums of different kind (one as a row sum and the other as a column sum or similar) as such sums have a common summand. If both $n$ and $-n$ arise as diagonal sums, $n$ must be even (otherwise the middle summand would be common) and each row and each diagonal must contain one $1$ and one $-1$. But then sums $n-1$

and $-(n-1)$ would be impossible to achieve. Hence $n$ and $-n$ must be either both row sums or both column sums. W.l.o.g., assume that they are both row sums.

The number $n-1$ can arise only as the sum of $n-1$ numbers $1$ and one number $0$. As $-1$ occurs in each column and each long diagonal, $n-1$ can be obtained as a row sum only. Similarly, $-(n-1)$ can be obtained as a row sum only. The number $n-2$ can arise as the sum of either $n-2$ numbers $1$ and two numbers $0$ or $n-1$ numbers $1$ and one number $-1$. As either two $-1$s or numbers $0$ and $-1$ occur in each column and each long diagonal, $n-2$ can be obtained as a row sum only. Similarly, $-(n-2)$ can be obtained as a row sum only.

Therefore, the table must contain at least $6$ rows.

An example of a $6 \times 6$ table that fulfils the conditions is shown in Fig. 42: the row sums from the top to the bottom are $6$, $5$, $-5$, $-4$, $4$, and $-6$, the column sums from the left to the right are $0$, $-2$, $1$, $2$, $-1$, $0$, and the diagonal sums are $3$ and $-3$.

Fig. 42