, 2023

The answer is all even $n$ and $m$. Clearly all cases of $2k \times 2h$ can be tiled using $k \times h$ copies of the following $2 \times 2$ baby array:

Now we prove that the condition of the problem can be satisfied only when both $n$ and $m$ are even. Let us number the columns in order as 1, 2, ..., $n$, and the rows in order as 1, 2, ..., $m$, and denote the tile in column $i$, row $j$ as $(i, j)$. Let us consider all the "odd" cells where both $i$ and $j$ are odd, and all the "even" cells where both $i$ and $j$ are even. Notice that regardless of the initial facing directions of the babies, after the teacher claps twice, all babies originally on odd cells will appear on even cells, all babies originally on even cells will appear on odd cells, and no baby originally on a cell that is neither odd nor even will appear on an odd or even cell. This means the number of odd cells and even cells must be equal (otherwise there will certainly be babies colliding). However, when $n$ and $m$ are not both even, the number of odd cells and even cells must be different. Therefore, the condition of the problem can be satisfied only when both $n$ and $m$ are even.