Find all integers $k$ with the following property: there exist infinitely many positive integers $n$ such that $(n+k) \nmid \mathrm{C}_{2 n}^{n}$.

保留了源文本的换行和格式。请注意，这里的 $\mathrm{C}_{2 n}^{n}$ 表示从 $2n$ 个不同元素中选取 $n$ 个元素的组合数，通常写作 $\binom{2n}{n}$。
【Analysis】Using the conclusion of Question 5, we know that $k=1$ does not satisfy the problem.
When $k=0$, take $n$ as an odd prime $p$.
Obviously, in base $p$
$$
p+p=(\overline{10})_{p}+(\overline{10})_{p}=(\overline{20})_{p}
$$

does not carry. By Kummer's theorem, we know $p \nmid \mathrm{C}_{2 p}^{p}$, i.e., $k=0$ satisfies the problem.
Similarly, for negative integer $k=-l$, take
$n+k=n-l=p($ prime $p>2 l)$.
Obviously, in base $p$
$$
(p+l)+(p+l)=(\overline{1 l})_{p}+(\overline{1} l)_{p}=(\overline{2 L})_{p}
$$

does not carry $(L=2 l)$.
By Kummer's theorem, we know $p \nmid \mathrm{C}_{2(p+l)}^{p+l}$.
Therefore, negative integer $k$ satisfies the problem.
When $k=2$, take
$$
n+2=3^{j}, n+2=(\overline{1 \underbrace{00 \cdots 0}_{j \uparrow}})_{3}\left(j \in \mathbf{Z}_{+}\right) \text {. }
$$

Obviously, in base 3, the first $j-1$ digits of $n$ are 2, and the last digit is 1, i.e., $n+n$ carries $j-1$ times.

By Kummer's theorem, we know $3^{j-1} \| \mathrm{C}_{2 n}^{n}$, i.e., $k=2$ satisfies the problem.

Similarly, when $k \geqslant 3$, take $n+k=p^{j}$ (prime $p \in(k, 2 k)$ (Bertrand-Chebyshev theorem), $\left.j \in \mathbf{Z}_{+}\right)$.

Obviously, in base $p$, the first $j-1$ digits of $n$ are $p-1$, and the last digit is $p-k \in(0, k)$, i.e., $n+n$ carries $j-1$ times.

By Kummer's theorem, we know $p^{j-1} \| \mathrm{C}_{2 n}^{n}$, i.e., $k \geqslant 3$ satisfies the problem.
In summary, $k \neq 1$.
【Note】When $k \geqslant 3$, letting $n+k$ be a positive integer power of 2 also works.