Chess sum symmetry for which natural numbers? The entries of a $n\times n$ chess table are black and white. For which natural numbers $n$, we can fill the table entries with numbers $\{1,2,...,n\}$ where:

*

*In each row, we have all ${1,2,...,n}$, and the sum of black entries are equal to white entries.

*In each column, we have all ${1,2,...,n}$, and the sum of black entries are equal to white entries.

My attempt: $1+2+...+n$ should be even, so $4|n(n+1)$.
 A: We can build such a table if and only if $\mathbf{4 | n}$.
Proving $4 | n$ works
Here's a diagram that helped me see what's going on here:

As that diagram shows, it can be useful to break the chessboard down into 4 colors instead of 2. (Then just think of red/blue cells as secretly white and green/yellow cells as black.) Notice how the red cells form a (spaced out) $\frac n 2 \times \frac n 2$ grid, and likewise for the other colors. Requiring the sum of whites vs blacks to be equal in each row/column is the same as requiring that the row/column sums of the $\frac n 2 \times \frac n 2$ single-color matrices are all equal.
We assumed $4 | n$, so we can divide the numbers $\{1, 2, \cdots, n\}$ into two piles $A$ and $B$ such that $|A| = |B| = \frac n 2$ and $\text{sum}(A) = \text{sum}(B) = \frac{n(n+1)}{4}$. Now fill the red $\frac n 2 \times \frac n 2$ submatrix with the elements of $A$ in any Latin square configuration, and do the same with the blue $\frac n 2 \times \frac n 2$ submatrix. Next, fill the green and yellow submatrices with a Latin square using $B$'s elements. The resulting matrix satisfies all required conditions.
Proving other values of $n$ don't work
As already stated in the question, $n \equiv 1 \text{ or } 2 \mod 4$ are doomed from the start because $\frac{n(n+1)}{2}$ will be odd. So we only need to worry about $n \equiv 3 \mod 4$.
We reuse the same "4 colors" idea from the proof above. This time $n$ is odd, so the 4 colored subgrids will have different sizes:

Now, let's compute the total value across all red+yellow cells in two different ways.
First: Every column in the grid has a sum of $\frac{n(n+1)}{2}$. There are $\frac{n+1}{2}$ red/yellow columns, so the total of all numbers in red and yellow spaces is $\left(\frac{n(n+1)}{2} \right) \left( \frac{n+1}{2}\right) = \frac{n(n+1)^2}{4}$.
Second: In each row, the red or yellow cells must contribute exactly half of the row total, or $\frac{n(n+1)}{4}$. There are $n$ rows total, so the sum over all red and yellow spaces is $\frac{n^2 (n+1)}{4}$.
At this point we've derived two conflicting formulas for the sum across all red and yellow squares, so we conclude it's not possible to satisfy the listed requirements in this case.
