Clearly true if n is a multiple of 2 or 3. Also, you can show it's true for n prime using Fermat's little theorem. You can also show that if n divides 3{a+b} - 2^{a+b} and n divides 3^a - 2^a then n divides 3^b - 2^b.
Not sure where to go from there though, or if I should be trying something completely different...
Take $p$ to be the smallest prime divisor of $n$. You have that $p$ divides $3^{p-1}-2^{p-1}$ and also $3^n-2^n$. So $p$ divides $3^{\operatorname{gcd}(p-1,n)}-2^{\operatorname{gcd}(p-1,n)}$. However it is easy to see that this gcd must equal $1$, so $p$ divides $3-2$, and we obtain the desired contradiction.
EDIT: I correct my answer. We have $3^n -2^n\equiv 1 \mod p$ for the smallest prime divisor $p$ of $n$. The proof is the same as the one by Gjergji Zaimi. Write $n=p^k\ell$ and use Fermat's little theorem repeatedly: $$ 3^n-2^n\equiv 3^{\ell}-2^{\ell}\equiv 3^{(p-1,\ell)}-2^{(p-1,\ell)}\equiv 1\mod p, $$ since $(p-1,\ell)=1$, because every prime divisor of $\ell$ is bigger than $p-1$.
