Given a positive integer $n$. Find the largest integer $M$, such that for any positive integers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
{\left[\sqrt{a_{1}}\right]+\left[\sqrt{a_{2}}\right]+\cdots+\left[\sqrt{a_{n}}\right]} \\
\geqslant\left[\sqrt{a_{1}+a_{2}+\cdots+a_{n}+\operatorname{Min}\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}}\right],
\end{array}
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
II. The maximum value of the integer $M$ is $\left[\frac{n(n-1)}{3}\right]$.
First, prove: $M \leqslant\left[\frac{n(n-1)}{3}\right]$.
Take $a_{i}=3$ $(i=1,2, \cdots, n)$, then
$$
\begin{array}{l}
n \geqslant[\sqrt{3 n+3 M}] \\
\Rightarrow \sqrt{3 n+3 M}\sqrt{a_{1}+a_{2}+\cdots+a_{n}+\frac{1}{3} n(n-1) a_{1}} .
\end{array}
$$

Square both sides of the above equation, and note that $a_{i} \leqslant k_{i}^{2}+2 k_{i}$ $(i=1,2, \cdots, n)$, thus it suffices to prove:
$2 \sum_{1 \leqslant i<j \leqslant n} k_{i} k_{j} \geqslant \frac{1}{3} n(n-1)\left(k_{1}^{2}+2 k_{1}\right)$.
In fact,
$$
\begin{array}{l}
\sum_{1 \leqslant i<j \leqslant n} k_{i} k_{j}=k_{1} \sum_{i=2}^{n} k_{i}+\sum_{2 \leqslant i<j \leqslant n} k_{i} k_{j} \\
\geqslant k_{1}(n-1) k_{1}+\frac{1}{2}(n-1)(n-2) k_{1}^{2} \\
=\frac{1}{2} n(n-1) k_{1}^{2} \geqslant \frac{1}{6} n(n-1)\left(k_{1}^{2}+k_{1}\right) .
\end{array}
$$

Therefore, the original problem is proved.