Find all positive integers $a_1, a_2, \ldots, a_n$ 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 $(a_{k+1}-1)a_{k-1} \geq a_k^2(a_k - 1)$ for $k = 1,2,\ldots,n-1$.
1. **Base Case Verification:**
   We start by verifying the base case for \(i = n-1\). We need to show that:
   \[
   \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{n-2}}{a_{n-1}} = \frac{a_{n-1}}{a_n} \leq \frac{a_{n-1}}{a_n - 1}
   \]
   This inequality simplifies to:
   \[
   \frac{a_{n-1}}{a_n} \leq \frac{a_{n-1}}{a_n - 1}
   \]
   Since \(a_n > 1\), the inequality holds because \(\frac{a_{n-1}}{a_n} \leq \frac{a_{n-1}}{a_n - 1}\) is always true.

2. **Induction Hypothesis:**
   Assume that for some \(m < n\), the following holds:
   \[
   \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{m-1}}{a_m} \leq \frac{a_m}{a_{m+1} - 1}
   \]
   We need to show that:
   \[
   \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{m-2}}{a_{m-1}} \leq \frac{a_{m-1}}{a_m - 1}
   \]

3. **Inductive Step:**
   It suffices to show that:
   \[
   \frac{a_{m-1}}{a_m} \leq \frac{a_{m-1}}{a_m - 1} - \frac{a_m}{a_{m+1} - 1}
   \]
   Expanding and simplifying, we get:
   \[
   \frac{a_{m-1}}{a_m} \leq \frac{a_{m-1}(a_m - 1) - a_m(a_{m+1} - 1)}{(a_m - 1)(a_{m+1} - 1)}
   \]
   This reduces to:
   \[
   (a_{m+1} - 1)a_{m-1} \geq a_m^2(a_m - 1)
   \]
   This inequality is given to be true by the problem statement.

4. **Conclusion from Induction:**
   By induction, we have shown that:
   \[
   \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{i-1}}{a_i} \leq \frac{a_i}{a_{i+1} - 1}
   \]
   for all \(i < n\).

5. **Determining \(a_i\):**
   Since:
   \[
   \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{i-1}}{a_i} = \frac{a_i}{a_{i+1}} + \frac{a_{i+1}}{a_{i+2}} + \cdots + \frac{a_{n-1}}{a_n} \geq \frac{a_i}{a_{i+1}}
   \]
   we have:
   \[
   \frac{a_i}{a_{i+1}} \leq \frac{99}{100} - \frac{a_0}{a_1} - \frac{a_1}{a_2} - \cdots - \frac{a_{i-1}}{a_i} \leq \frac{a_i}{a_{i+1} - 1}
   \]

6. **Finding Specific Values:**
   - For \(i = 0\):
     \[
     \frac{a_0}{a_1} = \frac{1}{a_1} \leq \frac{99}{100}
     \]
     This implies \(a_1 = 2\).
   - For \(i = 1\):
     \[
     \frac{a_1}{a_2} = \frac{2}{a_2} \leq \frac{99}{100} - \frac{1}{2} = \frac{49}{100}
     \]
     This implies \(a_2 = 5\).
   - Continuing this process, we find:
     \[
     a_3 = 56, \quad a_4 = 25^2 \cdot 56
     \]

The final answer is \( \boxed{ (a_0, a_1, a_2, a_3, a_4) = (1, 2, 5, 56, 25^2 \cdot 56) } \).