Nostalgia

If an arrangement realizes $X=M$, transposing it realizes $X=-M$. It is therefore enough to construct nonnegative $X$.

The column-maximum sum is maximized by making $N^2-N+1,\ldots,N^2$ the maxima of the $N$ columns. If the row maxima in increasing order are $M_1<\cdots<M_N$, the first $i$ rows contain $Ni$ values, so necessarily $M_i\geq Ni$. This yields the exact absolute bound

$$L=\frac{N(N-1)^2}{2}.$$

The row-major arrangement attains $X=L$.

For $0\leq X\leq(N-1)^2$, start from an $X=0$ arrangement with $N^2-N+1,\ldots,N^2$ on the diagonal. Lower only the first-row maximum to $N^2-N+1-X$ and move the displaced large value elsewhere in the first column. This changes $S_1-S_2$ by exactly $X$.

For larger $X$, start from the maximum row-major construction. Keep the last row fixed so the column maxima stay fixed, and increase the maxima of the first $N-1$ rows from the bottom upward until their sum has increased by $L-X$. The target maxima satisfy $M_i\geq Ni$ and remain strictly increasing, allowing the remaining smaller values to be distributed among the rows.

For negative $X$, transpose the constructed matrix. The construction takes $O(N^2)$ time. Solution written by GPT5.6