Find the smallest positive integer $n$, such that $x_{1}{ }^{3}+x_{2}{ }^{3}+\cdots+x_{n}{ }^{3}=2002^{2002}$ has integer solutions.
Analysis We first estimate the value of $n$, then construct a solution.
Solution $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002=667 \times 3+1$,
so $2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9)$.
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $\quad x \in \mathbf{Z}$.
Thus, $x_{1}{ }^{3}, x_{1}{ }^{3}+x_{2}{ }^{3}, x_{1}{ }^{3}+x_{2}{ }^{3}+x_{3}{ }^{3} \not=4(\bmod 9)$.
Since $2002=10^{3}+10^{3}+1+1$, then
$$
\begin{aligned}
2002^{2002} & =2002 \times\left(2002^{667}\right)^{3} \\
& =\left(10 \times 2002^{657}\right)^{3}+\left(10 \times 2002^{667}\right)^{3}+\left(2002^{667}\right)^{3}+\left(2002^{667}\right)^{3} .
\end{aligned}
$$

Therefore, $n=4$.