For natural numbers $m$ and $n$, an $m\times n$ grid of squares can be tiled with tiles of the form
completely filling the grid, without overlapping, if and only if $m,n\geq2$ and $6\mid mn$.
It is clear that the conditions are necessary, but is there a nice way to see that they are sufficient? By 'nice' I mean at least without distinguishing cases for when $2\nmid m$, $3\nmid m$ or both.
If one direction is divisible by $3$ and the other direction is divisible by $2$ it is obvious that $3\times 2$ rectangles will do the trick.
If one direction $m$ is divisible by $6$ it is easy to create both $2 \times 6$ and $3\times 6$ blocks. And then any integer $n\ge 2$ is expressible in the form $n=2a+3b$ for positive integers $a, b$. Allocate $a$ columns of width $2$ and $b$ columns of width $3$.
I think you have to distinguish these two cases in some way. Anyway, that was the simplest I could think of.