2. Find all positive integers $n$ such that for any positive integers $a, b, c$ satisfying $a+b+c \mid a^{2}+b^{2}+c^{2}$, we have $a+b+c \mid a^{n}+b^{n}+c^{n}$.
(Li Changyong)
2. The positive integers $n=3 k-1$ and $n=3 k-2\left(k \in \mathbf{N}^{*}\right)$ that satisfy the requirements.

Let $S_{n}=a^{n}+b^{n}+c^{n}$, then by Newton's identities (or by Vieta's formulas and the characteristic root method), we have
$$
S_{n+3}=(a+b+c) S_{n+2}-(a b+b c+c a) S_{n+1}+a b c S_{n} .
$$

The condition is $S_{1} \mid S_{2}$, which implies $a+b+c \mid(a+b+c)^{2}-2(a b+b c+c a)$, i.e., $a+b+c \mid 2(a b+b c+c a)$.
(1) If $a+b+c$ is odd, then $S_{n}$ is always odd, so $a+b+c \mid a b+b c+c a$. Thus, $a+b+c \mid S_{n+3} \Leftrightarrow a+b+c \mid a b c S_{n}$. Therefore, if $a+b+c \mid S_{n}$, then $a+b+c \mid S_{n+3}$ must hold.
(2) If $a+b+c$ is even, then $S_{n}$ is always even, so $a+b+c \mid(a b+b c+c a) S_{n+1}$. Thus, $a+b+c \mid S_{n+3} \Leftrightarrow a+b+c \mid a b c S_{n}$. Therefore, if $a+b+c \mid S_{n}$, then $a+b+c \mid S_{n+3}$ must hold.

From the condition, we know that $n=1$ and $n=2$ satisfy the requirements, so $n=3 k-2$ and $n=3 k-1\left(k \in \mathbf{N}^{*}\right)$ also satisfy the requirements.

When $n=3 k\left(k \in \mathbf{N}^{*}\right)$, take $a=1, b=2, c=11$. It is easy to verify that $a+b+c \mid a^{2}+b^{2}+c^{2}$ holds. On the other hand, it is easy to see that $1^{3 k}+2^{3 k}+11^{3 k} \equiv 3(\bmod 7)$, i.e., $7 \nmid 1^{3 k}+2^{3 k}+11^{3 k}$, hence $14 \nmid 1^{3 k}+2^{3 k}+11^{3 k}$, i.e., $a+b+c \nmid a^{3 k}+b^{3 k}+c^{3 k}$. Therefore, $n=3 k$ does not satisfy the requirements.
In conclusion, the positive integers $n=3 k-1$ and $n=3 k-2\left(k \in \mathbf{N}^{*}\right)$ that satisfy the requirements.