Question 4 Let $P$ be a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0 .
\end{array}
$$
$$
\text { If } P(3 n+1)=730 \text {, find } n \text {. }
$$