6. 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 $a_{1}<a_{2}<\cdots<a_{n}$.
6. When $a_{i}=i, i=1,2, \cdots, n$,
$$
\lambda \leqslant(n-2) \div \frac{n-1}{2}=\frac{2 n-4}{n-1} \text {. }
$$

The following proves the inequality
$$
a_{n}^{2} \geqslant \frac{2 n-4}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n}
$$

for any integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $0<a_{1}<a_{2}<\cdots<a_{n}$.

Since $a_{k} \leqslant a_{n}-(n-k), k=1,2, \cdots, n-1$. $a_{n} \geqslant n$, so,
$$
\begin{array}{l}
\frac{2 n-4}{n-1} \sum_{k=1}^{n-1} a_{k} \leqslant \frac{2 n-4}{n-1}\left[(n-1) a_{n}-\frac{n(n-1)}{2}\right] \\
\leqslant(2 n-4) a_{n}-n(n-2) \\
=(n-2)\left(2 a_{n}-n\right) \leqslant\left(a_{n}-2\right) a_{n} .
\end{array}
$$

Thus, $a_{n}^{2} \geqslant \frac{2 n-4}{n-1}\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+2 a_{n}$.
Therefore, the maximum value of $\lambda$ is $\frac{2 n-4}{n-1}$.