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. 

Define $k_n$ to be the least $k$ such that every triangle can be further triangulated into $nk_n$ small triangles with $k_n$ congruence classes. Equivalently, into $n$ pairwise congruent sets (in the sense of the question) each of size $k_n.$ Then certainly any $m$-gon can be partitioned into $(m-2)$ triangles and then into $n$ pairwise congruent sets each of size $(m-2)k_n.$ It seems plausible that this is optimal, though justifying that with a proof might be a challenge.

- We know that $k_n \le n.$ 
- $k_{t^2}=1$
- If $n=st^2$ with $s$ square free, then $k_n \le k_s$ 

It seems plausible that $k_{st^2}=k_s,$ but, again, that is not a proof.

However it seems most fruitful to concentrate on $k_s$ For $s$ square free.



- $k_2=2$
- $k_3=2$ (split into kites then bisect each).
- $k_{ab} \leq k_ak_b$ since we can partition each of $k_a$ types into $k_b$ subtypes. In particular:
- $k_{3s} \le 2k_s$ 

I suspect that $k_p=p$ for prime $p \neq 2.$ If so, then the last result is the only interesting use of the previous result.