Consider the top-left cell, it follows that the sum of the numbers in the first row is zero. Considering the cells in the first row and summing them up, it follows that the sum of the numbers in the table is zero. Analogously, it follows that the sum of the numbers in the first column is zero. Using this and writing the relation for the top-right cell, it follows that the top-right cell is zero. We claim that the sum of the four vertices of any rectangle is zero. To see this, consider all cells in the two columns corresponding to the vertices of the rectangle. Add up the resulting relations and use the fact that the sum of the cells in the table is zero. Now, if the numbers in the first row are indexed from right to left as $a_1 = 0, a_2, \dots, a_n = 1$, and the numbers in the first column are indexed from top to bottom as $b_1, \dots, b_n = 1$, we have: $x_{ij}$ (cell $i$ row, $j$ column)

$$x_{ij} = -a_i - b_j$$

By writing the relation for the cells in the first row, we get:
$$(s + 1)a_{n-s} = b_2 + \cdots + b_{2+s}$$
which implies:
$$a_{n-s} \le \max(|b_2|, \dots, |b_{2+s}|)$$
Similarly,
$$|b_{n-s}| \le \max(|a_2|, \dots, |b_{2+s}|)$$
Consider the cell with the maximum absolute value in the row and column. By symmetry, we must have $a_{n-s} = |b_{2+s}| = \dots = |b_2|$. Considering $b_2$, all $a_i$ must have the same sign, and all $b_i$ must have the same sign. By writing the relation for the first row, we have $a_i = -n/2$, $b_i = 1/n$.