2. Find the conditions that the sequence of positive integers $\left\{a_{0}, a_{1}, \ldots, a_{n+1}\right\}$ must satisfy so that for any positive integer $n$, $\frac{a_{0}}{a_{1}}+\frac{a_{0}}{a_{2}}+\ldots+\frac{a_{0}}{a_{n}}+\frac{2017}{a_{n+1}}=1$ holds. (2017 is a prime number)
2, Solution: The condition is equivalent to $\left\{\begin{array}{c}\frac{a_{0}}{a_{1}}+\frac{2017}{a_{2}}=1 \\ \frac{a_{0}}{a_{n+1}}+\frac{2017}{a_{n+2}}=\frac{2017}{a_{n+1}}\end{array}\right.$, First, use mathematical induction to prove: $2017^{n-2} \mid a_{n}, n \geq 2$. When $n=2$, the conclusion is obviously true;

When $n=3$, since $\frac{a_{0}}{a_{2}}+\frac{2017}{a_{3}}=\frac{2017}{a_{2}}$, rearranging gives $a_{0} a_{3}=2017\left(a_{3}-a_{2}\right)$, clearly $a_{3}-a_{2}>0$, so $2017 \mid a_{0} a_{3}$; if $2017 \mid a_{0}$, then $a_{3} \mid\left(a_{3}-a_{2}\right)$, but $a_{3}>a_{3}-a_{2}>0$, which is a contradiction. Hence, $2017 \mid a_{3}$, the conclusion holds.
Assume that when $n=m(m \geq 3)$, the conclusion holds. From $\frac{a_{0}}{a_{m}}+\frac{2017}{a_{m+1}}=\frac{2017}{a_{m}}$, rearranging gives: $a_{0} a_{m+1}=2017\left(a_{m+1}-a_{m}\right)$.

Assume $2017^{\alpha} \| a_{m+1}$, noting that 2017 does not divide $a_{0}$, and $2017^{m-2} \mid a_{m}$, let $v_{p}(n)$ be the highest power of the prime $p$ in the standard factorization of the positive integer $n$, then we know $v_{2017}\left(a_{0} a_{m+1}\right)=\alpha$, and $v_{2017}\left(2017\left(a_{m+1}-a_{m}\right)\right) \geq \min \{\alpha, m-2\}+1$, hence $\alpha \geq \min \{\alpha, m-2\}+1$, i.e., $\alpha \geq m-1$. The conclusion holds. According to induction, $2017^{n-2} \mid a_{n}, n \geq 2$ holds.
Let $a_{n}=2017^{n-2} b_{n}(n \geq 2), b_{n} \in Z^{+}$, then for any positive integer $n \geq 2$, we have
$$
\frac{a_{0}}{a_{n}}+\frac{2017}{a_{n+1}}=\frac{2017}{a_{n}} \Rightarrow \frac{a_{0}}{2017^{n-2} b_{n}}+\frac{2017}{2017^{n-1} b_{n+1}}=\frac{2017}{2017^{n-2} b_{n}} \Rightarrow \frac{a_{0}}{b_{n}}+\frac{1}{b_{n+1}}=\frac{2017}{b_{n}} \Rightarrow b_{n+1}=\frac{b_{n}}{2017-a_{0}}
$$

Thus, for any positive integer $n \geq 2$, we have $b_{n}=\frac{b_{2}}{\left(2017-a_{0}\right)^{n-2}}$. Since $b_{n} \in Z^{+}$, it follows that $2017-a_{0}=1$, then $b_{n}=b_{2}=a_{2}, a_{0}=2016$. Since $\frac{a_{0}}{a_{1}}+\frac{2017}{a_{2}}=1$, i.e., it is sufficient for $a_{1}, a_{2}$ to satisfy $\frac{2016}{a_{1}}+\frac{2017}{a_{2}}=1$. In fact, let $a_{0}=2016, a_{n}=2017^{n-2} a_{2}(n \geq 2)$, for any $\left(a_{1}, a_{2}\right)$ satisfying $\frac{2016}{a_{1}}+\frac{2017}{a_{2}}=1$ form a sequence that meets the conditions.