The answer is $n = 4$. We call those two kinds of nonsquare tiles ducks.
First, the left-hand-side figure and the middle figure below show that $n = 4$ works.

Second, we show that $n$ must be even. We mark the (infinite) chessboard with $\times$ in the pattern indicated in the right-hand-side figure. It is easy to see that each $2 \times 2$ covers exactly an even number of crosses (either two or four crosses) and each duck covers exactly an odd number of crosses (either one or three crosses). It follows that we must have an even number of ducks in $R$, i.e. $n$ is even.

Third, we show that $n \ge 2$. If $n = 2$, then $R$ can be tiled by two $2 \times 2$ squares. It is clear that these two squares much share a common edge. Hence, we can have only two possibilities: