9. Find the least positive integral value of $n$ for which the equation
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$
has integer solutions $\left(x_{1}, x_{2}, x_{3}, \cdots, x_{n}\right)$.
9. Answer: 4
Solution. Since $2002 \equiv 4(\bmod 9) \quad 4^{3} \equiv 1(\bmod 9)$ and $2002=667 \times 3+1$, it follows that $2002^{2002} \equiv 4^{667 \times 3+1} \equiv 4(\bmod 9)$. Observe that for positive integers $x$, the possible residues modulo 9 for $x^{3}$ are $0 \pm 1$. Therefore, none of the following
$$
x_{1}^{3} x_{1}^{3}+x_{2}^{3} \quad x_{1}^{3}+x_{2}^{3}+x_{3}^{3}
$$
can have a residue of 4 modulo 9 . However, since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, it follows that
$$
\begin{aligned}
2002^{2002} & =2002 \cdot\left(2002^{667}\right)^{3} \\
& =\left(10^{3}+10^{3}+1^{3}+1^{3}\right)\left(2002^{667}\right)^{3} \\
& =\left(10 \cdot 2002^{667}\right)^{3}+\left(10 \cdot 2002^{667}\right)^{3}+\left(2002^{667}\right)^{3}+\left(2002^{667}\right)^{3}
\end{aligned}
$$

This shows that $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{4}=2002^{2002}$ is indeed solvable. Hence the least integral value of $n$ is 4 .