The 23rd question: Given a positive real number $k(k>2)$ and a positive integer $n(n \geq 3)$, find the largest positive real number $\lambda$, such that if the positive real numbers $\mathrm{a}_{\mathrm{i}}(1 \leq \mathrm{i} \leq \mathrm{n})$ satisfy $\left(\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{i}}\right)\left(\sum_{\mathrm{i}=1}^{\mathrm{n}} \frac{1}{a}\right)<\lambda$, then it must be true that $\mathrm{a}_{1}+\mathrm{a}_{2}<k \mathrm{a}_{3}$.
The 23rd question:
Solution: The required $\lambda_{\max }=\left(\sqrt{\mathrm{k}+\frac{4}{\mathrm{k}}+5}+\mathrm{n}-3\right)^{2}$.

If $\lambda>\left(\sqrt{\mathrm{k}+\frac{4}{\mathrm{k}}+5}+\mathrm{n}-3\right)$, we take $\mathrm{a}_{1}=\mathrm{a}_{2}=\mathrm{k}, \mathrm{a}_{3}=2, \mathrm{a}_{4}=\mathrm{a}_{5}=\ldots=\mathrm{a}_{\mathrm{n}}=$ $2 \sqrt{\frac{k(k+1)}{k+4}}$, 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) \geq\left(\sum_{i=1}^{n} a_{i}\right)\left(\frac{4}{a_{1}+a_{2}}+\sum_{i=3}^{n} \frac{1}{a_{i}}\right) \\
\geq\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}
\end{array}
$

Thus, we know
$
\begin{array}{l}
\left(a_{1}+a_{2}+a_{3}\right)\left(\frac{4}{a_{1}+a_{2}}+\frac{1}{a_{3}}\right)<k+\frac{4}{k}+5 \\
\Rightarrow\left(\frac{a_{1}+a_{2}}{a_{3}}-k\right)\left(\frac{a_{1}+a_{2}}{a_{3}}-\frac{4}{k}\right)<0 \\
\Rightarrow \frac{4}{k}<\frac{a_{1}+a_{2}}{a_{3}}<k \\
\Rightarrow a_{1}+a_{2}<k a_{3}
\end{array}
$

In summary, the maximum value of the required $\lambda$ is $\left(\sqrt{\mathrm{k}+\frac{4}{\mathrm{k}}+5}+\mathrm{n}-3\right)^{2}$.