Find all positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$\frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}$$

where $a_{0}=1$, and $\left(a_{k+1}-1\right) a_{k-1} \geqslant a_{k}^{2}-1(k=1,2, \cdots, n-1)$.
4. Clearly $a_{k}>a_{k-1}$, and $a_{k} \geqslant 2 . k=1,2, \cdots, n-1$.

From the condition, we have
$$\frac{a_{k-1}}{a_{k}} \leqslant \frac{a_{k-1}}{a_{k}-1}-\frac{a_{k}}{a_{k+1}-1}$$

Summing the above inequality for $k=i+1, i+2, \cdots, n$, we get
$$\frac{a_{i}}{a_{i+1}}+\frac{a_{i+1}}{a_{i+2}}+\cdots+\frac{a_{n-1}}{a_{n}}<\frac{a_{i}}{a_{i+1}-1} .$$

When $i=0$, by the problem statement and the above inequality, $\frac{1}{a_{1}} \leqslant \frac{99}{100}<\frac{1}{a_{1}-1}$. Then
$$\frac{100}{99} \leqslant a_{1}<\frac{100}{99}+1 \text {, hence } a_{1}=2 \text {. }$$

When $i=1, 2, 3$, similarly, we can find $a_{2}=5, a_{3}=56, a_{4}=78400$.
Therefore, $\frac{1}{a_{5}} \leqslant \frac{1}{a_{4}}\left(\frac{99}{100}-\frac{1}{2}-\frac{2}{5}-\frac{5}{56}-\frac{56}{25-56^{2}}\right)=0$, so there is no $a_{5}$, thus, the positive integers sought are $a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=78400$.