Question 4 Let $P$ be a polynomial of degree $3n$, such that
$$
\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2, \\
P(1)=P(4)=\cdots=P(3 n-2)=1, \\
P(2)=P(5)=\cdots=P(3 n-1)=0 .
\end{array}
$$
$$
\text { If } P(3 n+1)=730 \text {, find } n \text {. }
$$
Solving, we know
$$
\sum_{k=0}^{3 n+1}(-1)^{k} \dot{\mathrm{C}}_{3 n+1}^{k} P(3 n+1-k)=0,
$$

which means $729+2 \sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1}+\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=0$
Using the multi-section formula, we can find
$$
\begin{array}{l}
\sum_{j=0}^{n}(-1)^{3 j+1} \mathrm{C}_{3 n+1}^{3 j+1} \\
=-2(\sqrt{3})^{3 n-1} \cos \frac{(2 n-1) \pi}{6}, \\
\sum_{j=0}^{n}(-1)^{3 j} \mathrm{C}_{3 n+1}^{3 j}=2(\sqrt{3})^{3 n-1} \cos \frac{(3 n+1) \pi}{6} .
\end{array}
$$

Thus, $729-4(\sqrt{3})^{3 n-1} \cos \frac{(3 n-1) \pi}{6}+$
$$
2(\sqrt{3})^{3 n-1} \cos \frac{(3 n+1) \pi}{6}=0 \text {. }
$$

By distinguishing the parity of $n$, it is easy to find that only $n=4$.