Every $n$ is possible. 

As you point out, it suffices to answer the question for triangles. You can divide a triangle $T$ into $n^2$ congruent triangles  similar to $T.$ Then these can be partitioned into $n$ sets of $n$ which are congruent as sets and, indeed, have all members congruent.

It is interesting to note that, If you triangulate an $m$-gon $P$ into $k \geq m-2$ triangles and follow this procedure you get $kn^2$ triangles in $k$ congruence classes. These can be assembled into $n^2$ congruent $m$-gons similar to  $P$ and triangulated in the same way.