Coins on a polygon

[Phlembac Adib Hasan]

A number of coins have been placed at each vertex of the regular $n$-gon $A_1 A_2 . . . A_n$ . These coins may be re-arranged using the following move: two coins may be chosen, and each moved to an adjacent vertex, subject to the requirement that one must be moved clockwise and the other anti-clockwise. (Thus, for example, you may move a coin from each of $A_1$ and $A_5$ to vertices $A_2$ and $A_4$ respectively: each coin ends up on a vertex adjacent to the one it started on, and they move in opposite directions.)

Suppose that there are initially $k$ coins at vertex $A_k$ for each $k$, $1 \leq k \leq n$. For which $n$ is it possible to re-arrange the coins using finitely many such moves so that there are exactly $n + 1 − k$ coins at vertex $A_k$ , $1 \leq k \leq n$?

Note:
This problem has an almost instant invariant solution. The key idea is old and appears as early as in ISL 1994 C5. So if you want more of a challenge, try to transform it into a pure algebra problem.

[Nirjhor]

Simply assigning the weight $j$ to the coins at $A_j$ reveals $n=6x\pm 1$.