2. Given that $n$ is a positive integer, such that there exist positive integers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)=100 n .
$$

Find the maximum possible value of $n$.
(Lin Jin, problem contributor)
2. The maximum possible value of $n$ is 9702.

Obviously, from the given equation, we have $\sum_{i=1}^{n} x_{i} \geqslant n$. Therefore, $\prod_{i=1}^{n} x_{i} \leqslant 100$.
Since equality cannot hold, then $\prod_{i=1}^{n} x_{i} \leqslant 99$.
$$
\begin{array}{l}
\text { and } \prod_{i=1}^{n} x_{i}=\prod_{i=1}^{n}\left(\left(x_{i}-1\right)+1\right) \\
\geqslant \sum_{i=1}^{n}\left(x_{i}-1\right)+1=\sum_{i=1}^{n} x_{i}-n+1, \\
\text { then } \sum_{i=1}^{n} x_{i} \leqslant \prod_{i=1}^{n} x_{i}+n-1 \leqslant n+98 \\
\Rightarrow 99(n+98) \geqslant 100 n \\
\Rightarrow n \leqslant 99 \times 98=9702 .
\end{array}
$$

Taking $x_{1}=99, x_{2}=x_{3}=\cdots=x_{9702}=1$, can make the equality in (1) hold.