Let $ a_1, a_2, \ldots, a_n$ denote number of times we re-distribute $ 2$ stones from a point $ P_n$, with $ a_{n+i} = a_i$. Then to restore to initial conditions, we require $ 2a_k = a_{k-1}+a_{k+1}$. Summing that with $ 2a_{k+1}=a_k+a_{k+2}$ yields $ a_{k+2} - a_{k+1} = a_k - a_{k-1} = m$ for some integer $ m$. For odd $ n$, it follows that $ a_k - a_{k-1} = m$ for all $ k$. Then, $ 0 = (a_2 - a_1) + (a_3 - a_2) + \cdots + (a_1 - a_n) = mn \Longrightarrow m = 0$, so all of the $ a_i$ are equal. For even $ n$, we have $ a_{2k} - a_{2k-1} = m_1, a_{2k+1} - a_{2k} = m_2$, with $ 0 = \frac n2 \cdot m_1 + \frac n2 \cdot m_2 \Longrightarrow m_1 = -m_2$. Substituting into $ 0 = 2a_{2k} - a_{2k-1} + a_{2k+1} = m_1 - m_2 = 2m_1$, and again all of the $ a_i$ are equal. Clearly $ n | N = \sum a_i$ then.