5. Find all positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that
$$
\begin{array}{l}
\quad \frac{99}{100}=\frac{a_{0}}{a_{1}}+\frac{a_{1}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}, \\
\text { where } a_{0}=1,\left(a_{k+1}-1\right) a_{k-1} \geqslant a_{k}^{2}\left(a_{k}-1\right), k=1,2, \cdots, \\
n-1 .
\end{array}
$$
Solution: Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers satisfying the given conditions.
Since $a_{0}=1$, we have $a_{1} \neq 1$. Thus, $\frac{a_{0}}{a_{1}}=1>\frac{99}{100}$. Therefore, $a_{1} \geqslant 2$, and $a_{1}>a_{0}$. Assume $a_{k}>a_{k-1}, a_{k} \geqslant 2$. Then
$$
a_{k-1} \geqslant \frac{a_{i}^{2}\left(a_{k}-1\right)}{a_{k-1}}+1>\frac{a_{i}^{2}\left(a_{k}-1\right)}{a_{k-1}} \geqslant \frac{a_{i}^{2}}{a_{h-1}^{2}}>a_{k} .
$$

From the above, we have
$a_{k+1}>a_{k}, a_{k+1} \geqslant 2$, where $k=0,1,2, \cdots, n-1$. Rewriting the inequality $\left(a_{k+1}-1\right) a_{k-1} \geqslant a_{k}^{2}\left(a_{k}-1\right)$ as
$$
\frac{a_{k-1}}{a_{k}\left(a_{k}-1\right)} \geqslant \frac{a_{k}}{a_{k+1}-1} \text {, }
$$

we get $\frac{a_{k-1}}{a_{i}} \leqslant \frac{a_{k-1}}{a_{k}-1}-\frac{a_{k}}{a_{k-1}-1}$. For $k=i+1, i+2, \cdots, n-1$, and $\frac{a_{n-1}}{a_{n}}<\frac{a_{n-1}}{a_{n}-1}$, summing up we get
$$
\frac{a_{i}}{a_{i, 1}}+\frac{a_{t+1}}{a_{i+2}}+\cdots+\frac{a_{n-1}}{a_{n}}<\frac{a_{i}}{a_{i+1}-1} .
$$

When $i=0$, we have
$$
\frac{1}{a_{1}} \leqslant \frac{99}{100}<\frac{1}{a_{1}-1},
$$

which implies $\frac{100}{99} \leqslant a_{1}<\frac{199}{99}$, so $a_{1}=2$.
When $i=1$, we have
$$
\frac{a_{1}}{a_{2}} \leqslant \frac{99}{100}-\frac{1}{a_{1}}<\frac{a_{1}}{a_{2}-1} .
$$

which implies $\frac{200}{49} \leqslant a_{2}<\frac{200}{49}+1$, so $a_{2}=5$.
When $i=2$, we have
$$
\frac{1}{a_{3}} \leqslant \frac{1}{a_{2}}\left(\frac{99}{100}-\frac{1}{a_{1}}-\frac{a_{1}}{a_{2}}\right)<\frac{1}{a_{3}-1},
$$

which implies $55 \frac{5}{9} \leqslant a_{3}<55 \frac{5}{9}+1$. So $a_{3}=56$.
When $i=3$, we have
$$
\frac{1}{a_{4}} \leqslant \frac{1}{a_{3}}\left(\frac{99}{100}-\frac{1}{2}-\frac{2}{5}-\frac{5}{56}\right)<\frac{1}{a_{4}-1} .
$$

which implies $56 \times 14 \times 100 \leqslant a_{4}<56 \times 14 \times 100+1$, so
$$
a_{4}=78400 \text {. }
$$

When $i=4$, we have
$$
\frac{1}{a_{5}} \leqslant \frac{1}{a_{4}}\left(\frac{99}{100}-\frac{1}{2}-\frac{2}{5}-\frac{5}{56}-\frac{56}{78400}\right)=0,
$$

which is impossible.
Therefore, $a_{1}=2, a_{2}=5, a_{3}=56, a_{4}=78400$ is the unique solution.