Let $n$ have $k$ digits. Then we want to have $2 \cdot 10^{k+1}+10n+1 = 33n$. Or in other words: $2 \cdot 10^{k+1}+1 = 23n$. So lets find $k$ with $2 \cdot 10^{k+1} \equiv -1 \mod 23$ or then $10^{k+1} \equiv 11 \mod 23$. $k=2$ works, giving $n=87$.