Example 9 For any positive integer $n$, let $x_{n}=\mathrm{C}_{2 n}^{n}$. Find all positive integers $h(h>1)$, such that the sequence of remainders of $\left\{x_{n}\right\}$ modulo $h$ is eventually periodic.
【Analysis】 $h=2$ is the solution we are looking for.
If $h$ is not a power of 2, let $p$ be a prime factor of $h$, and $p$ is odd. Then there exist positive integers $N$ and $T$ such that the sequence $\left\{x_{n}\right\}$ modulo $p$ has a periodic remainder sequence after the $N$-th term with period $T$.
Take a positive integer $k$ such that $p^{k}>\max \{N, T\}$.
Consider $n=(p-1) p^{k}+r\left(r=0,1, \cdots, p^{k}-1\right)$, then $n+n$ will carry over in the first digit in base $p$.
By Kummer's Theorem, we know $p \mid \mathrm{C}_{2 n}^{n}$.
At this point, the number of $n$ taken is greater than the period $T$.
Therefore, when $n \geqslant N$, $p \mid \mathrm{C}_{2 n}^{n}$ always holds.
But when $n=p^{k}$, $n+n$ does not carry over in base $p$.
By Kummer's Theorem, we know $p \nmid \mathrm{C}_{2 p^{k}}^{p^{k}}$, leading to a contradiction.
Thus, $h$ can only be a power of 2, let $h=2^{u}\left(u \in \mathbf{Z}_{+}\right)$.
If $u \geqslant 2$, similarly, there exist positive integers $N$ and $T$ such that the sequence $\left\{x_{n}\right\}$ modulo $2^{u}$ has a periodic remainder sequence after the $N$-th term with period $T$.
Take a positive integer $k$ such that $2^{k}>\max \{N, T\}$.
Consider $n=2^{k+u-1}+\cdots+2^{k}+r\left(r=0,1, \cdots, 2^{k}-1\right)$, then $n+n$ will carry over at least $u$ times in binary.
By Kummer's Theorem, we know $2^{u} \mid C_{2 n}^{n}$. At this point, the number of $n$ taken is greater than the period $T$. Therefore, when $n \geqslant N$, $2^{u} \mid \mathrm{C}_{2 n}^{n}$ always holds.
But when $n=2^{k}$, $n+n$ only carries over once in binary.
By Kummer's Theorem, we know $2 \| \mathrm{C}_{2^{k+1}}^{2 k}$, leading to a contradiction.
If $u=1$, i.e., $h=2$, then $\mathrm{C}_{2 n}^{n}=2 \mathrm{C}_{2 n-1}^{n-1}$, so $\left\{x_{n}\right\}$ modulo 2 is always 0, which meets the requirement.
【Comment】When analyzing the power of a binomial coefficient, generally consider Kummer's Theorem. In this problem, the periodic sequence is a condition, and the feature "the period length is finite" provides a breakthrough for the proof by contradiction.