1. Find the smallest positive integer $n$, such that
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$

has integer solutions.
(Uzbekistan provided)
Solution: Since $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002$ $=667 \times 3+1$, therefore,
$$
2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9) \text {. }
$$

Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $x$ is an integer, thus,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) \text {. }
$$

Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l} 
2002^{2002}=2002 \times\left(2002^{667}\right)^{3} \\
=\left(10 \times 2002^{667}\right)^{3}+\left(10 \times 2002^{667}\right)^{3}+ \\
\left(2002^{667}\right)^{3}+\left(2002^{667}\right)^{3} .
\end{array}
$$

Therefore, $n=4$.