Solution: $a_{i} \in \mathbf{N}^{+}$, there are nine classes modulo 9. It is not difficult to verify that the corresponding $a_{i}^{3}$ has only three cases: $a_{i}^{3} \equiv 0,1,-1(\bmod 9)$.
And $(2002)^{2005} \equiv 4^{2005}=\left(4^{3}\right)^{668} \times 4 \equiv 1^{668} \times 4=4(\bmod 9)$.
When $n=1,2,3$, $a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3} \not \equiv 4(\bmod 9)$, so $n \geqslant 4$.
$(2002)^{2005}=\left[(2002)^{668}\right]^{3} \times 2002=\left[(2002)^{668}\right]^{3} \times\left[10^{3}+10^{3}+\right.$
$\left.1^{3}+1^{3}\right]=\left[10 \times(2002)^{668}\right]^{3} \times 2+\left[(2002)^{668}\right]^{3} \times 2$, so the minimum value of $n$ is 4.
Example 3.1.2 $a_{i} \in \mathbf{N}^{+}$, find the minimum value of $n$ that satisfies $a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=(2002)^{2005}$.