Croatian Mathematical Olympiad

Since $3n-2 \mid n^{3n-2}-3n+1$, it follows that $3n-2 \mid n^{3n-2}-1$, hence $n^{3n-2} \equiv 1 \pmod{3n-2}$.

Note that $n$ must be odd (otherwise the even $3n-2$ would divide the odd $n^{3n-2}-3n+1$, which is impossible).

Let $p > 2$ be the smallest prime factor of the odd $3n-2$. Obviously $p \nmid n$, hence by Fermat's little theorem we have $n^{p-1} \equiv 1 \pmod{p}$. We also have $n^{3n-2} \equiv 1 \pmod{p}$.

Let $r$ be the order of $n$ modulo $p$. Then $r \mid p-1$ and $r \mid 3n-2$. Since $r \mid \gcd(p-1, 3n-2)$ and $\gcd(p-1, 3n-2) = 1$, it must be $r=1$, i.e. $n \equiv 1 \pmod{p}$.

Considering $p \mid n-1$ and $p \mid 3n-2$, it follows that $p \mid 1$, which contradicts the assumption about the existence of prime factor of $3n-2$. Hence $3n-2=1$, i.e. $n=1$, and that is indeed the only solution.