8. If the sum of the volumes of $n$ cubes with side lengths as positive integers is $2002^{2005}$. Find the minimum value of $n$.


8. Solution: Let the side lengths of $n$ cubes be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2005}$.
$$
\begin{array}{l}
\because 2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), \\
\therefore 2002^{2005} \equiv 4^{2005} \equiv 4(\bmod 9), \\
\because x^{3} \equiv 0, \pm 1(\bmod 9), x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \neq \\
4(\bmod 9), \therefore n \geqslant 4, \\
\because 2002^{2005}=2002^{2004}\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
\quad=\left(2002^{668}\right)^{3}\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
\quad=\left(2002^{668} \times 10\right)^{3}+\left(2002^{668} \times 10\right)^{3}+
\end{array}
$$
$$
\left(2002^{668}\right)^{3}+\left(2002^{668}\right)^{3}, \therefore n_{\text {min}}=4 .
$$