$\square$ Example 5 Given two positive integers $n \geqslant 2$ and $T \geqslant 2$, find all positive integers $a$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have $\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}<T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}$, where $S_{k}$ $=a_{1}+a_{2}+\cdots+a_{k}$. (1992 China National Training Team Selection Exam Problem)
$$\text { Solve } \begin{aligned}
& \sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}=\sum_{k=1}^{n} \frac{1}{S_{k}}\left[\left(k+\frac{a}{2}\right)^{2}-k^{2}\right] \\
= & \frac{1}{S_{1}}\left[\left(1+\frac{a}{2}\right)^{2}-1^{2}\right]+\frac{1}{S_{2}}\left[\left(2+\frac{a}{2}\right)^{2}-2^{2}\right] \\
& +\frac{1}{S_{3}}\left[\left(3+\frac{a}{2}\right)^{2}-3^{2}\right]+\cdots+\frac{1}{S_{n}}\left[\left(n+\frac{a}{2}\right)^{2}-n^{2}\right] \\
= & \frac{1}{S_{1}}\left(1+\frac{a}{2}\right)^{2}+\left[\frac{1}{S_{2}}\left(2+\frac{a}{2}\right)^{2}-\frac{1}{S_{1}}\right]+\left[\frac{1}{S_{3}}\left(3+\frac{a}{2}\right)^{2}-\frac{1}{S_{2}} 2^{2}\right] \\
& +\cdots+\left[\frac{1}{S_{n}}\left(n+\frac{a}{2}\right)^{2}-\frac{1}{S_{n-1}}(n-1)^{2}\right]-\frac{n^{2}}{S_{n}} \\
& =\frac{1}{S_{1}}\left(1+\frac{a}{2}\right)^{2}-\frac{n^{2}}{S_{n}}+\sum_{k=2}^{n}\left[\frac{1}{S_{k}}\left(k+\frac{a}{2}\right)^{2}-\frac{1}{S_{k-1}}(k-1)^{2}\right] .
\end{aligned}$$

For $k=2,3, \cdots, n$, let $a_{k} t_{k}=S_{k-1}, t_{k}>0$. Then, we have
$$\begin{aligned}
& \frac{1}{S_{k}}\left(k+\frac{a}{2}\right)^{2}-\frac{1}{S_{k-1}}(k-1)^{2}=\frac{\left(k+\frac{a}{2}\right)^{2}}{S_{k-1}+a_{k}}-\frac{1}{S_{k-1}}(k-1)^{2} \\
= & \frac{\left(k+\frac{a}{2}\right)^{2}}{a_{k}\left(t_{k}+1\right)}-\frac{(k-1)^{2}}{a_{k} t_{k}}=\frac{1}{a_{k} t_{k}\left(t_{k}+1\right)}\left[t_{k}\left(k+\frac{a}{2}\right)^{2}-(k-1)^{2}\left(t_{k}+1\right)\right] \\
= & \frac{1}{a_{k} t_{k}\left(t_{k}+1\right)}\left\{\left(\frac{a}{2}+1\right)^{2} t_{k}\left(t_{k}+1\right)-\left[\left(\frac{a}{2}+1\right) t_{k}-(k-1)\right]^{2}\right\} \\
\leqslant & \frac{(a+2)^{2}}{4 a_{k}}
\end{aligned}$$

From (1) and (2), we get
$$\begin{aligned}
& \sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}} \leqslant \frac{1}{S_{1}}\left(1+\frac{a}{2}\right)^{2}-\frac{n^{2}}{S_{n}}+\frac{(a+2)^{2}}{4} \sum_{k=2}^{n} \frac{1}{a_{k}} \\
= & \frac{(a+2)^{2}}{4} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}} .
\end{aligned}$$

When the positive integer $a$ satisfies $a \leqslant 2(T-1)$, by $\frac{(a+2)^{2}}{4} \leqslant T^{2}$, we know that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the inequality
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}} \leqslant T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}}2(T-1)$$ holds. When $a \geqslant 2 T-1$, there must exist positive numbers $a_{1}, a_{2}, \cdots, a_{n}$ such that the inequality in the problem does not hold.

For any given $a_{1}>0$, let $a_{k}=\frac{(a+2)}{2(k-1)} S_{k-1}(k=2,3, \cdots, n)$. Then, $a_{1}, a_{2}, \cdots, a_{n}$ are uniquely determined, where $a_{k}$ is determined to make $\frac{a+2}{2} t_{k}=k-1$, for $k=2,3, \cdots, n$. At this point, (2) becomes an equality. Using (1) and the equality case of (2), we have
$$\begin{aligned}
\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}} & =\frac{(a+2)^{2}}{4} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}} \\
& =\left[\frac{(a+2)^{2}}{4}-1\right] \sum_{k=1}^{n} \frac{1}{a_{k}}+\left(\sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}}\right) .
\end{aligned}$$

Using the AM-GM inequality, we know
$$S_{n} \sum_{k=1}^{n} \frac{1}{a_{k}} \geqslant n^{2}$$

From (3) and (4), we get
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}} \geqslant\left[\frac{(a+2)^{2}}{4}-1\right] \sum_{k=1}^{n} \frac{1}{a_{k}}$$

Since the positive integer $a \geqslant 2 T-1$, we know
$$a+2 \geqslant 2 T+1, \frac{(a+2)^{2}}{4}-1 \geqslant \frac{(2 T+1)^{2}}{4}-1>T^{2}$$

Thus, from (5), we get
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}>T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}$$

In summary, the positive integers $a$ that satisfy the problem's requirements are $1,2,3, \cdots, 2(T-1)$.