7. Given a positive integer $n(n \geqslant 2)$, find the largest real number $\lambda$, such that the inequality $a_{n}^{2} \geqslant \lambda\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+$ $2 a_{n}$, holds for any positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}<a_{2}<\cdots<a_{n}$.
7. First, derive the range of $\lambda$ for a special sequence $\left\{a_{n}\right\}$, then provide a proof for any sequence $\left\{a_{n}\right\}$.

Let $a_{i}=i(i=1,2, \cdots, n)$.
From the inequality, it is easy to find that $\lambda \leqslant \frac{2(n-2)}{n-1}$.
We now prove: For any positive integers $a_{1}<a_{2}<\cdots<a_{n}$, we have
$$
a_{n}^{2} \geqslant \frac{2(n-2)}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n} \text {. }
$$

By $a_{k}+(n-k) \leqslant a_{n}$, i.e., $a_{k} \leqslant a_{n}-(n-k)(k=1,2, \cdots, n-1)$, and $a_{n} \geqslant n$, we have
$$
\begin{array}{l}
\frac{2(n-2)}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \leqslant \frac{2(n-2)}{n-1}\left\{(n-1) a_{n}-[(n-1)+(n-2)+\cdots+2+1]\right\} \\
=\frac{2(n-2)}{n-1}\left[(n-1) a_{n}-\frac{1}{2} n(n-1)\right] \\
=2(n-2) a_{n}-n(n-2)=(n-2)\left(2 a_{n}-n\right) . \\
\text { And }\left(a_{n}^{2}-2 a_{n}\right)-(n-2)\left(2 a_{n}-n\right)=a_{n}\left[\left(a_{n}-n\right)+(2-n)\right]+n(n-2) \\
=(n-2)\left(n-a_{n}\right)+a_{n}\left(a_{n}-n\right)=\left(a_{n}-n\right)\left[a_{n}-(n-2)\right] \geqslant 0,
\end{array}
$$

Thus, inequality $(*)$ holds.
Therefore, the maximum value of $\lambda$ is $\frac{2(n-2)}{n-2}(n \geqslant 2)$.