Given a positive number $k(k>2)$ and a positive integer $n(n \geqslant 3)$. Find the maximum positive number $\lambda$, such that if positive numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)<\lambda,
$$

then it must be true that $a_{1}+a_{2}<k a_{3}$.
$$
\begin{array}{l}
\text { II. } \lambda_{\max }=\left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2} . \\
\text { If } \lambda>\left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2}, \text { let } \\
a_{1}=a_{2}=k, a_{3}=2, \\
a_{4}=a_{5}=\cdots=a_{n}=\sqrt{\frac{4 k(k+1)}{k+4}} . \\
\text { Then }\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) \\
=\left(\sqrt{k+\frac{4}{k}+5}+n-3\right)^{2}\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) \\
\geqslant\left(\sum_{i=1}^{n} a_{i}\right)\left(\frac{4}{a_{1}+a_{2}}+\sum_{i=3}^{n} \frac{1}{a_{i}}\right) \\
\geqslant\left(\sqrt{\left(a_{1}+a_{2}+a_{3}\right)\left(\frac{4}{a_{1}+a_{2}}+\frac{1}{a_{3}}\right)}+n-3\right)^{2} . \\
\text { Therefore, } k+\frac{4}{k}+5>\left(a_{1}+a_{2}+a_{3}\right)\left(\frac{4}{a_{1}+a_{2}}+\frac{1}{a_{3}}\right) . \\
\text { Let } \frac{a_{1}+a_{2}}{a_{3}}=x . \text { Then } \\
(x-k)\left(x-\frac{4}{k}\right)<0 \Rightarrow \frac{4}{k}<x<k \\
\Rightarrow a_{1}+a_{2}<k a_{3} .
\end{array}
$$