Answer

The answer is all $\boxed{n \ge 2}$. Let $A_i = \{m \in \mathbb{N} \mid \nu_2(m - 1) = i - 1\}$ for all $1 \le i \le n - 1$ and $A_n = \{m \in \mathbb{N} \mid \nu_2(m - 1) \ge n - 1\}$, where $\nu_2(m)$ is the largest possible integer $k$ such that $2^k \mid m$. Note that $\nu_2(0) = \infty$, so $1 \in A_n$. It is easy to see $A_1, A_2, \dots, A_n$ is a partition of $\mathbb{N}$. We show $f(A_i) \subseteq A_i$ for all $1 \le i \le n$. If $a, b, c \in A_1 \Leftrightarrow a, b, c$ are even, then $abc - b - c + 2$ is even $\Leftrightarrow abc - b - c + 2 \in A_1$. So $f(A_1) \subseteq A_1$. For all $2 \le i \le n - 1$, if $a, b, c \in A_i$, then \[(abc - b - c + 2) - (a - 1) - 1 = (a - 1) (b - 1)(c - 1) + (a - 1) (b - 1) + (b - 1)(c - 1) + (c - 1)(a - 1) \equiv 0 \pmod{2^{2i-2}}.\]Since $2^i \mid 2^{2i-2}$, $(abc - b - c + 2) - 1 \equiv a - 1 \equiv 2^{i-1} \pmod{2^i}$, so $abc - b - c + 2 \in A_i$. Hence, $f(A_i) \subseteq A_i$. If $a, b, c \in A_n$, then $2^{n-1} \mid a - 1, b - 1, c - 1$, so \[2^{n-1} \mid (a - 1) (b - 1)(c - 1) + (a - 1) (b - 1) + (b - 1)(c - 1) + (c - 1)(a - 1) + (a - 1) = (abc - b - c + 2) - 1.\]So $abc - b - c + 2 - 1 \in A_n$, thus, $f(A_n) \subseteq A_n$. Therefore, $f(A_i) \subseteq A_i$ for all $1 \le i \le n$, and we are done. $\Box$