1) We prove that the statement is true for all odd prime $p$ instead of $2017$. Suppose there exists a positive integer $k$ such that $\frac{p^k}{2} < n < p^k$. We have
$$v_p(x_n) = v_p(C_{2n}^n) = v_p((2n)!) - 2v_p(n!).$$
Because $\frac{p^k}{2} < n < p^k$ so $p^k < 2n < 2p^k < p^{k+1}$, hence
$$v_p((2n)!) = \left\lfloor \frac{2n}{p} \right\rfloor + \left\lfloor \frac{2n}{p^2} \right\rfloor + \dots + \left\lfloor \frac{2n}{p^k} \right\rfloor.$$
On the other hand, for every $x \in \mathbb{R}$, we have $\lfloor 2x \rfloor \ge 2\lfloor x \rfloor$, the equality holds for $\{x\} < \frac{1}{2}$. Combined with the condition $\frac{p^k}{2} < n < p^k$, we have
$$v_p((2n)!) > 2 \left( \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \dots + \left\lfloor \frac{n}{p^k} \right\rfloor \right) = 2v_p(n!),$$
or $v_p(x_n) > 0$ so we conclude that $x_n$ is divisible by $p$.

2) Suppose $h > 1$ is the number satisfying the problem requirement. For every odd prime number $p$ then $p \mid h$, we have the remainders sequence $x_n$ modulo $p$ is also periodic. Using the result of part 1), for $\frac{p^k}{2} < n < p^k$ then
$$x_n \equiv 0 \pmod{p}.$$
Choose $k$ be big enough for $\frac{p^k}{2} > T+1$, we conclude that all the remainders of $x_n$ divide $p$ equal $0$ for all $n \ge n_0$, where $n_0 \in \mathbb{Z}^+$ is large enough. However, choose $t \in \mathbb{Z}^+$ which is large enough for $p^t - 1 > 2n_0$ and set $n = \frac{p^t - 1}{2}$ we have $v_p(x_n) = 0$, so $x_n$ is not divisible by $p$, absurd.

Therefore $h$ has only prime divisors of $2$ or $h = 2^k$ with $k$ is a positive integer. If $k > 1$, we choose $r = k-1$ and consider the number $n$ has the form $n = 2^{a_1} + \dots + 2^{a_r}$, where $a_1 > \max\{T, N\}$ where $T, N$ are constants in the hypothesis of $h$. Then
$$v_2(x_n) = 2S_2(n) - S_2(2n) = r,$$
where $S_2(x)$ is the sum of the digits in the binary representation. Hence $x_n \equiv 2^{k-1} \pmod{h}$. However, for every $i \in \mathbb{Z}^+$ that $i < 2^{a_1}$ then the $1$ digit in the binary representation of $n+i$ adds at least $1$ unit, so that $x_{n+i} \equiv 0 \pmod{h}$. Since $a_1 > \max\{T, N\}$ then $x_n \equiv x_{n+T} \equiv 0 \pmod{h}$, absurd.

So $k = 1$ and so $h = 2$. This is the answer of problem, because it is easy to see that $x_n$ is an even number for every positive integer $n$, as follows: if $n = 2^{a_1} + \dots + 2^{a_r}$ with $0 \le a_1 < \dots < a_r$ then
$$2n = 2^{a_1+1} + \dots + 2^{a_r+1}$$
so
$$v_2(x_n) = 2S_2(n) - S_2(2n) = r \ge 1.$$
Hence $x_n$ even. Therefore, $h = 2$ is the number we have to find.