6. Let $n(n \geqslant 2)$ be a given positive integer, find all integer tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy the conditions:
(1) $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$;
(2) $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1$.
6. From the conditions, we have
$$
\begin{array}{l}
\sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=\sum_{k=1}^{n} a_{k}^{2}-2 n \sum_{k=1}^{n} a_{k}+n^{3} \\
\leqslant\left(n^{3}+1\right)-2 n^{3}+n^{3}=1 . \\
\text { Therefore, } \sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=0 \text { or } 1 . \\
\text { If } \sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=0 \text {, then } a_{1}=a_{2}=\cdots=a_{n}=n . \\
\text { If } \sum_{k=1}^{n}\left(a_{k}-n\right)^{2}=1, \text { then }\left|a_{k}-n\right|(k=1,2, \cdots,
\end{array}
$$
$n$ ) has exactly one value of 1, and the remaining $n-1$ values are 0. Without loss of generality, assume $\left|a_{1}-n\right|=1, a_{2}=a_{3} \cdots=a_{n}=n$. In this case, combining $a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2}$, we know $a_{1}=n+1$, hence $\sum_{k=1}^{n} a_{k}^{2}=(n+1)^{2}+(n-1) n^{2}=n^{3}+2 n+1>n^{3}+1$. This is a contradiction.
In conclusion, only the set of numbers $(n, n, \cdots, n)$ satisfies the conditions.