Three. (20 points) Let $n(n \geqslant 2)$ be a given positive integer. Find all integer tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, satisfying the conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$;
(2) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$.
Three, by the mean inequality we have
$$
\begin{array}{l}
n=\frac{n^{2}}{n} \leqslant \frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \\
\leqslant \sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}} \leqslant \sqrt{\frac{n^{3}+1}{n}} \\
<n+\frac{1}{2 n^{2}},
\end{array}
$$
(Note: $\left(n+\frac{1}{2 n^{2}}\right)^{2}=\frac{n^{3}+1}{n}+\left(\frac{1}{2 n^{2}}\right)^{2}$ )
Thus, $\square$
$$
\begin{array}{l}
n^{2} \leqslant a_{1}+a_{2}+\cdots+a_{n} \\
<n^{2}+\frac{1}{2 n}<n^{2}+1 .
\end{array}
$$

Therefore, $a_{1}+a_{2}+\cdots+a_{n}=n^{2}$.
Hence, the equality in equation (1) holds.
This is true if and only if $a_{1}=a_{2}=\cdots=a_{n}=n$.
Thus, there is only one solution $(n, n, \cdots, n)$.