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 National IMO Training Team Selection Exam Problem)
$$\begin{array}{l}
\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}=\sum_{k=1}^{n} \frac{1}{S_{k}}\left[\left(k+\frac{a}{2}\right)^{2}-k^{2}\right]= \\
\frac{1}{S_{1}}\left[\left(1+\frac{a}{2}\right)^{2}-1^{2}\right]+\frac{1}{S_{2}}\left[\left(2+\frac{a}{2}\right)^{2}-2^{2}\right]+\frac{1}{S_{3}}\left[\left(3+\frac{a}{2}\right)^{2}-3^{2}\right]+\cdots+\frac{1}{S_{n}}[(n \\
\left.\left.+\frac{a}{2}\right)^{2}-n^{2}\right]= \\
\frac{1}{S_{1}}\left(1+\frac{a}{2}\right)^{2}+\left[\frac{1}{S_{2}}\left(2+\frac{a}{2}\right)^{2}-\frac{1}{S_{1}}\right]+\left[\frac{1}{S_{3}}\left(3+\frac{a}{2}\right)^{2}-\frac{1}{S_{2}} 2^{2}\right]+\cdots+\left[\frac{1}{S_{n}}(n+\right. \\
\left.\left.\frac{a}{2}\right)^{2}-\frac{1}{S_{n-1}}(n-1)^{2}\right]-\frac{n^{2}}{S_{n}}= \\
\frac{1}{S_{1}}\left(1+\frac{a}{2}\right)^{2}-\frac{n^{2}}{S_{n}}+\sum_{k=2}^{n}\left[\frac{1}{S_{k}}\left(k+\frac{a}{2}\right)^{2}-\frac{1}{S_{k-1}}(k-1)^{2}\right]
\end{array}$$

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

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

When the positive integer $a$ satisfies $a \leqslant 2(T-1)$, since
$$\frac{(a+2)^{2}}{4} \leq T^{2}$$

we know that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the inequality
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}} \leqslant T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}}$$

holds. When the positive integer $a \geqslant 2 \boldsymbol{T}^{-1}-1$, 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
$$a_{k}=\frac{(a+2)}{2(k-1)} S_{h-1},(k=2,3, \cdots, n)$$

Thus, $a_{1}, a_{2}, \cdots, a_{n}$ are uniquely determined, where $a_{k}$ is determined to make
$$\frac{a+2}{2} t_{k}=k-1$$

hold for $k=2,3, \cdots, n$: At this point, equation (2) becomes an equality. Using (1) and the equality case of (2), we have
$$\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}=\frac{(a+2)^{2}}{4} \sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}}=\left[\frac{(a+2)^{2}}{4}-1\right] \sum_{k=1}^{n} \frac{1}{a_{k}}+\left(\sum_{k=1}^{n} \frac{1}{a_{k}}-\frac{n^{2}}{S_{n}}\right)$$

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

From equations (8) and (9), 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 $a+2 \geqslant 2 T+1, \frac{(a+2)^{2}}{4}-1 \geqslant \frac{(2 T+F)^{2}}{4}$ $-1>T^{2}$,
thus, from equation (10) 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 $1 ; 2,3, \cdots, 2(T-1)$.