Sk.26 Too For a given prime $p$, find the number of different sequences of positive integers $a_{0}, a_{1}, a_{2}, \cdots$ that satisfy
$$
\frac{a_{0}}{a_{1}}+\frac{a_{0}}{a_{2}}+\cdots+\frac{a_{0}}{a_{n}}+\frac{p}{a_{n+1}}=1, n=1,2, \cdots
$$
Let $S_{n}-\frac{a_{0}}{a_{1}}+\frac{a_{0}}{a_{2}}+\cdots+\frac{a_{0}}{a_{n}}+\frac{p}{a_{n+1}}$. Then the sequence $\left\{S_{n}\right\}$ is a constant sequence if and only if for each $n \in \mathbf{N}^{*}, S_{n+1}-S_{n}=\frac{a_{0}}{a_{n+1}}+\frac{p}{a_{n+2}}-\frac{p}{a_{n+1}}=0$. That is, $a_{n+2}=\frac{p}{p-a_{0}} a_{n+1}$. Suppose the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies the given conditions, then for each $n(\geqslant 3)$, we have
$$
a_{n}=\frac{p}{p-a_{0}} a_{n-1}=\left(\frac{p}{p-a_{0}}\right)^{2} a_{n-2}=\cdots=\left(\frac{p}{p-a_{0}}\right)^{n-2} a_{2} \text {. }
$$

Since $a_{n} \in \mathbf{N}^{*},\left(p, p-a_{0}\right)=1$, for any positive integer $m,\left(p-a_{0}\right)^{m} \mid a_{2}$, thus $p-a_{0}=1$, i.e., $a_{0}=p-1$. Therefore,
$$
\frac{p-1}{a_{1}}+\frac{p}{a_{2}}=1, a_{n}=p^{n-2} a_{2},
$$

where $a_{1}, a_{2} \in \mathbf{N}^{*}, n=3,4, \cdots$.
Conversely, if the sequence $\left\{a_{n}\right\}$ satisfies (2), and $a_{0}=p-1$, then $S_{1}=1, S_{n+1}=S_{n}\left(n \in \mathbf{N}^{*}\right)$, thus the sequence $\left\{a_{n}\right\}$ satisfies the given conditions. Therefore, the problem is reduced to: for a given prime $p$, find the number of positive integer solutions $\left(a_{1}, a_{2}\right)$ to the equation
$$
\frac{p-1}{a_{1}}+\frac{p}{a_{2}}=1
$$

Transform the equation (3) into
$$
\left[a_{1}-(p-1)\right]\left(a_{2}-p\right)=p(p-1) .
$$

Since $p$ is a prime, $p$ cannot simultaneously divide $a_{1}-(p-1)$ and $a_{2}-p$ (otherwise $p^{2} \mid\left(a_{2}-\right.$ $p)\left[a_{1}-(p-1)\right]$, contradicting (4)), i.e., $p \mid a_{1}-(p-1)$ and $p \mid a_{2}-p$ are mutually exclusive.

Choose any positive divisor of $p-1$ as $a_{1}-(p-1)$, and let $a_{2}-p=k p$, where $k=\frac{p-1}{a_{1}-(p-1)}$, then we get one set of positive integer solutions to (3).

Choose any positive divisor of $p-1$ as $a_{2}-p$, and let $a_{1}-(p-1)=l p$, where $l=\frac{p-1}{a_{2}-p}$, then we get one set of positive integer solutions to (3).

Since the above two conditions are mutually exclusive, all the solutions obtained are different. Therefore, the number of positive integer solutions to (3) is twice the number of positive divisors of $p-1$, which is also the number of the required sequences.