Let $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, b_{2}, \cdots, b_{n} \in [1001,2002]$, and
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}$$
$$\begin{aligned}
& \text { Prove: } \frac{a_{1}^{3}}{b_{1}}+\frac{a_{2}^{3}}{b_{2}}+\cdots+\frac{a_{n}^{3}}{b_{n}} \leqslant \frac{17}{10}\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) . \\
& \frac{21 a_{m}^{2}-4 b_{m}^{2}}{10}-\frac{a_{m}^{3}}{b_{m}}=\left(2 a_{m}-b_{m}\right)\left(2 b_{m}-a_{m}\right)\left(\frac{a_{m}}{2 b_{m}}+\frac{1}{5}\right) \geqslant 0(0 \leqslant m \leqslant n) \\
\Rightarrow & \sum_{m=1}^{n} \frac{a_{m}^{3}}{b_{m}} \leqslant \frac{21}{10} \sum_{m=1}^{n} a_{m}^{2}-\frac{4}{10} \sum_{m=1}^{n} b_{m}^{2}=\frac{17}{10} \sum_{m=1}^{n} a_{m}^{2} .
\end{aligned}$$

Using the method of the user named myth on the mathlinks website, we have
$$\left(6 n-2+8\left[\frac{n-1}{2}\right]\right) \sum_{m=1}^{n} \frac{a_{m}^{3}}{b_{m}} \leqslant\left(13 n-9+8\left[\frac{n-1}{2}\right]\right) \sum_{m=1}^{n} a_{m}^{2}$$

When $n=9$, $\sum_{m=1}^{9} \frac{a_{m}^{3}}{b_{m}} \leqslant \frac{5}{3} \sum_{m=1}^{9} a_{m}^{2}$.
When $a_{1}=a_{2}=\cdots=a_{5}=b_{5}=\cdots=b_{9}=1001$,
$a_{6}=\cdots=a_{9}=b_{1}=\cdots=b_{4}=2002$, the equality holds.