58-Let $n(n \geqslant 2)$ be a given positive integer, find all integer solutions $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfying the conditions:
$$\begin{array}{l}
(-1) a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2} \\
(-2)=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1
\end{array}$$
(-2002 Western Mathematical Olympiad Problem)
58. Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be an integer solution satisfying the conditions, then by the Cauchy-Schwarz inequality we have
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \geqslant n^{3}$$

Combining this with $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$, we can conclude that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=n^{3}$ or $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$.

When $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=n^{3}$, by the equality condition of the Cauchy-Schwarz inequality, we have $a_{1}=a_{2}=\cdots=a_{n}$, i.e., $a_{i}^{2}=n^{2}, 1 \leqslant i \leqslant n$, and by $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$, we have $a_{1}=a_{2}=\cdots=a_{n}=n$.

When $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=n^{3}+1$, let $b_{i}=a_{i}-n$, then
$$\begin{aligned}
b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}= & \left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)-2 n\left(a_{1}+a_{2}+\cdots+a_{n}\right)+n^{3} \leq \\
& 2 n^{3}+1-2 n\left(a_{1}+a_{2}+\cdots+a_{n}\right) \leqslant 1
\end{aligned}$$

Thus, $b_{i}^{2}$ can only be 0 or 1, and at most one of $b_{1}^{2}, b_{2}^{2}, \cdots, b_{n}^{2}$ can be 1. If all are 0, then $a_{i}=n, a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=n^{3} \neq n^{3}+1$, which is a contradiction. If one of $b_{1}^{2}, b_{2}^{2}, \cdots, b_{n}^{2}$ is 1, then $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=n^{3} \pm 2 n+1 \neq n^{3}+1$, which is also a contradiction. Therefore, the only integer solution is $\left(a_{1}, a_{2}, \cdots, a_{n}\right)=$ $(n, n, \cdots, n)$.