Croatian Mathematical Olympiad

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

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 $n^{p-1}\equiv 1(\mathrm{mod}p)$. We also have $n^{3n-2}\equiv 1(\mathrm{mod}p)$.

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. $n\equiv 1(\mathrm{mod}p)$.

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.