4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2002^{2000}$, find the minimum value of $n$ and explain the reason.
4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.

Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\end{array}
$$

thus $x \neq 9$;
$$
\begin{array}{l}
2005-10^{3}-10^{3}=5,2005-10^{3}-9^{3}=276, \\
2005-10^{3}-8^{3}=493,2005-10^{3}-7^{3}-7^{3}>0,
\end{array}
$$

thus $x \neq 10$;
$2005-11^{3}-9^{3}0$,
thus $x \neq 11$;
$2005-12^{3}-7^{3}0$,
thus $x \neq 12$.
Therefore, $n=3$ is not possible.
In conclusion, $n_{\min }=4$.
(2) Let the edge lengths of $n$ cubes be $x_{1}, x_{2}, \cdots, x_{n}$, then
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2005} \text {. }
$$

By $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9)$, we get
$$
\begin{array}{l}
2002^{2005} \equiv 4^{2005} \equiv 4^{668 \times 3+1} \\
\equiv\left(4^{3}\right)^{668} \times 4 \equiv 4(\bmod 9)
\end{array}
$$

Also, when $x \in \mathbf{N}_{+}$, $x^{3} \equiv 0, \pm 1(\bmod 9)$, so,
$$
\begin{array}{l}
x_{1}^{3} \not \equiv 4(\bmod 9), x_{1}^{3}+x_{2}^{3} \not \equiv 4(\bmod 9), \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \not \equiv 4(\bmod 9) .
\end{array}
$$

Taking equation (1) modulo 9, and from equations (2) and (3), we know $n \geqslant 4$.
Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l} 
2002^{2005}=2002^{2004} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668}\right)^{3} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668} \times 10\right)^{3}+\left(2002^{668} \times 10\right)^{3}+ \\
\left(2002^{668}\right)^{3}+\left(2002^{668}\right)^{3} .
\end{array}
$$

Therefore, $n=4$ is the required minimum value.