Firstly, $n \equiv 1 \pmod{3}$ because otherwise one of them is divisible by $3$. Now, $n=1$ is a solution. Now, the whole idea of this problem is that when $n = 3k+1$, we have $37|111\cdots111$ with number of $1$'s divisible by 3. More formal: $37 \cdot 10^{\text{blabla}} + 7 \cdot \frac{10^{n-1}-1}{9}$ is divisible by $37$ as $10^{3k} \equiv (1000)^k \equiv 1^k \equiv 1 \pmod{37}$ so $37|10^{3k}-1$ done