Solution:

We shall prove that one can obtain the opposite table by rearranging the rows and columns of the initial table. Denote the columns from left to right and the rows from up to down by $1,2, \ldots, n$. Denote by $a_{ij}$ the number written in the $i$-th row and $j$-th column.

Exchanging rows and columns one obtains: $a_{11}=1$, $a_{12}=-1$, $a_{22}=1$, $a_{23}=-1$ (when $a_{21}=-1$ the assertion follows by induction using $2 \times 2$ and $(n-1) \times (n-2)$ tables), $a_{33}=1$, $a_{34}=-1$ (if $a_{31}=-1$, the assertion follows by induction using $3 \times 3$ and $(n-3) \times (n-3)$ tables) and so on.

It remains to prove the assertion for the table

Proceeding in the same way one obtains from the initial table the following one

$(B)$

Now applying the same moves for obtaining the table $B$ from the table $A$ but in reverse order one obtains the opposite of the initial table.