Example 11 Given two natural numbers $n$ and $m$ greater than 1, find all natural numbers $l$ such that for any positive numbers $a_{1}$, $a_{2}, \cdots, a_{n}$, we have
$$
\sum_{k=1}^{n} \frac{1}{s_{k}}\left(l k+\frac{1}{4} l^{2}\right)<m^{2} \sum_{k=1}^{n} \frac{1}{a_{k}},
$$

where $s_{k}=a_{1}+a_{2}+\cdots+a_{k}$.
(1992 China National Team Selection Test)
It is known that
\[
\begin{aligned}
\sum_{k=1}^{n} \frac{1}{s_{k}}\left(l k+\frac{1}{4} l^{2}\right) & =\sum_{k=1}^{n}\left[\frac{1}{s_{k}}\left(\frac{l}{2}+k\right)^{2}-\frac{k^{2}}{s_{k}}\right] \\
& =\left(\frac{l}{2}+1\right)^{2} \frac{1}{s_{1}}-\frac{n^{2}}{s_{n}}+\sum_{k=2}^{n}\left[\frac{1}{s_{k}}\left(\frac{l}{2}+k\right)^{2}-\frac{(k-1)^{2}}{s_{k-1}}\right] .
\end{aligned}
\]

Since for $k=2, \cdots, n$, we have
\[
\begin{aligned}
& \frac{1}{s_{k}}\left(\frac{l}{2}+k\right)^{2}-\frac{(k-1)^{2}}{s_{k-1}} \\
= & \frac{1}{s_{k} s_{k-1}}\left[\left(\frac{l}{2}+1\right)^{2} s_{k-1}+(l+2)(k-1) s_{k-1}+(k-1)^{2}\left(s_{k-1}-s_{k}\right)\right] \\
= & \frac{1}{s_{k} s_{k-1}}\left[\left(\frac{l}{2}+1\right)^{2} s_{k-1}-\left(\sqrt{a_{k}}(k-1)-\left(\frac{l}{2}+1\right) \frac{s_{k-1}}{\sqrt{a_{k}}}\right)^{2}+\left(\frac{l}{2}+1\right)^{2} \frac{s_{k-1}^{2}}{a_{k}}\right] \\
\leqslant & \frac{1}{s_{k} s_{k-1}}\left(\frac{l}{2}+1\right)^{2}\left(s_{k-1}+\frac{s_{k-1}^{2}}{a_{k}}\right)=\left(\frac{l}{2}+1\right)^{2} \frac{1}{a_{k}},
\end{aligned}
\]

Therefore, \(\sum_{k=1}^{n} \frac{1}{s_{k}}\left(l k+\frac{1}{4} l^{2}\right) \leqslant\left(\frac{l}{2}+1\right)^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{s_{n}}2(m-1)\). When \(l \geqslant 2 m-1\), it can be proven that there exist positive numbers \(a_{1}, a_{2}, \cdots, a_{n}\) such that the required inequality does not hold. In fact, for any given \(a_{1}>0\), let
\[
a_{k}=\frac{l+2}{2(k-1)} s_{k-1}, k=2,3, \cdots, n .
\]

From the previous proof, we know
\[
\sum_{k=1}^{n} \frac{1}{s_{k}}\left(l k+\frac{1}{4} l^{2}\right)=\left(\frac{l}{2}+1\right)^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{s_{n}}=\left[\left(\frac{l}{2}+1\right)^{2}-1\right] \sum_{k=1}^{n} \frac{1}{a_{k}}+\sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{s_{n}} \text{. }
\]

From \(l \geqslant 2 m-1\), we can deduce
\[
\left(\frac{l}{2}+1\right)^{2}-1 \geqslant\left(m+\frac{1}{2}\right)^{2}-1=m^{2}+m+\frac{1}{4}-1>m^{2} \text{. }
\]

Using the Cauchy-Schwarz inequality, we get
\[
n^{2} \leqslant\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{k=1}^{n} \frac{1}{a_{k}}\right)=s_{n} \cdot \sum_{k=1}^{n} \frac{1}{a_{k}},
\]
i.e., \(\sum_{k=1}^{n} \frac{1}{a_{k}}=\frac{n^{2}}{s_{k}} \geqslant 0\).
Thus, \(\sum_{k=1}^{n} \frac{1}{s_{k}}\left(l k+\frac{1}{4} l^{2}\right)>m^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}\).
Therefore, \(1,2, \cdots, 2(m-1)\) are all the natural numbers \(l\) that satisfy the requirement.