6 Let $n$ 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) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$.
(Supplied by Pan Chengbiao)
Let $\left(a_{1}, \cdots, a_{n}\right)$ be an integer solution, then by the Cauchy-Schwarz inequality, we have
$$
a_{1}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+\cdots+a_{n}\right)^{2} \geqslant n^{3} .
$$

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

If the former is true, then by the equality condition of the Cauchy-Schwarz inequality, we have $a_{1}=\cdots=a_{n}$, which requires $a_{i}^{2}=n^{2}, 1 \leqslant i \leqslant n$. Combining this with $a_{1}+\cdots+a_{n} \geqslant n^{2}$, we can only have $a_{1}=\cdots=a_{n}=n$.
If the latter is true, then let $b_{i}=a_{i}-n$, we have
$$
\begin{aligned}
b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2} & =\sum_{i=1}^{n} a_{i}^{2}-2 n \sum_{i=1}^{n} a_{i}+n^{3} \\
& =2 n^{3}+1-2 n \sum_{i=1}^{n} a_{i} \leqslant 1 .
\end{aligned}
$$

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 $b_{1}^{2}, \cdots, b_{n}^{2}$ are all zero, then $a_{i}=n, \sum_{i=1}^{n} a_{i}^{2}=n^{3} \neq n^{3}+1$, which is a contradiction; if one of $b_{1}^{2}, \cdots, b_{n}^{2}$ is 1, then $\sum_{i=1}^{n} a_{i}^{2}=n^{3} \pm 2 n+1 \neq n^{3}+1$, which is also a contradiction.
In summary, the only solution is $\left(a_{1}, \cdots, a_{n}\right)=(n, \cdots, n)$.