Problem 1480

Solution:
We will show that the answer is all pairs $(a,b)$ where either $a$ or $b$ (or both) is a multiple of $4$. First, partition the grid into four subgrids $A$, $B$, $C$, and $D$

Note that, for any given square $s$, all the square centres that are exactly $2$ away from the centre of $s$ are in the same partition as $s$ itself. This means that $s$ must be paired with some square which is in the same partition as itself. Hence, each partition of squares must have an even number of squares. Meanwhile, if each partition of squares has an even number of squares, one of the dimensions of the subgrid must be even, and hence we can pair up the squares along that even dimension.

Hence, it suffices to find all values of $a$ and $b$ which create four subgrids which all have an even number of squares. Consider $a$ modulo $4$: if $a \equiv 0$ (mod $4$), then all subgrids will have an even dimension along the axis of $a$. Otherwise, at least one of the subgrids will have an odd dimension along the axis of $a$. The same reasoning holds for $b$. Hence, if neither $a$ nor $b$ are multiples of $4$, then one of the partitions will have an odd number of squares. However, if either $a$ or $b$ is a multiple of $4$, then all partitions will have an even number of squares. Hence, we have shown that the answer is all pairs $(a,b)$ where either $a$ or $b$ (or both) is a multiple of $4$.