[Solution] It does not exist. If there exists $a \in(0,1)$ and an infinite sequence of positive numbers $\left\{a_{n}\right\}$ satisfying the conditions in the problem, then
$$a_{n} \geqslant \frac{n}{n+a}\left(1+a_{n+1}\right)>\frac{n}{n+1}\left(1+\dot{a}_{n+1}\right), n=1,2,3 \cdots$$

Thus, for any natural number $n$, we have
$$\begin{aligned} a_{1} & >\frac{1}{2}\left(1+a_{2}\right)=\frac{1}{2}+\frac{1}{2} a_{2}>\frac{1}{2}+\frac{1}{2} \cdot \frac{2}{3}\left(1+a_{3}\right) \\ & =\frac{1}{2}+\frac{1}{3}+\frac{1}{3} a_{3}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(1+a_{4}\right) \\ & >\cdots>\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n} a_{n} . \end{aligned}$$

Since $\lim _{n \rightarrow \infty}\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)=+\infty$,
this leads to a contradiction!
$$x_{n}=x_{1}, n=1,2, \cdots$$

That is, every term of $\left\{x_{n}\right\}$ is a non-zero integer. Conversely, if $\left\{x_{n}\right\}$ has infinitely many terms as integers, it is clear from (1) that $x_{1}=x_{2}$, and further, $x_{1}=x_{2}$ is a non-zero integer.

In summary, the necessary and sufficient condition for the non-zero sequence $\left\{x_{n}\right\}$ to have infinitely many terms as integers is that $x_{1}=$ $x_{2}$ is a non-zero integer.