Determine all integers $k$ such that there exist infinitely many positive integers $n$ not satisfying
$$n + k \mid \binom{2n}{n}.$$

We claim that all integers $k \neq 1$ satisfy the desired property.

First, recall that $\frac{1}{n + 1} \binom{2n}{n}$ is the $n$-th Catalan number. Since the Catalan numbers are a sequence of integers, it follows that $n + 1 \mid \binom{2n}{n}$ for all $n$. Hence, $k = 1$ certainly cannot satisfy the problem statement.

Now, we consider two cases:

**Case 1: $k \neq 2$.**

Suppose that $p$ is a prime divisor of $k$ and let $n = p^\alpha$ for any $\alpha \in \mathbb{N}$. Then, since $p \mid n + k$, in order to prove that $n + k \nmid \binom{2n}{n}$, it suffices to show that
$$p \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}.$$
Note that the greatest power of $p$ that divides any term in the numerator or denominator of $\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}$ is less than $p^{\alpha}$. Since the sets $\{1, 2, \cdots, n - 1\}$ and $\{n + 1, n + 2, \cdots, 2n - 1\}$ are congruent modulo $p^{\alpha}$, the numerator and denominator of the fraction $\frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}$ both contain the same number of factors of $p$. Therefore, $p \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)}$.

Now, if we can show that $p \nmid 2$, we will be able to conclude that $p \nmid \binom{2n}{n}$, as desired. Indeed, if $p \neq 2$, then trivially $p \nmid 2$. Meanwhile, if $p = 2$, then let us take $\alpha \geq 2$ so that $2^2 \mid n + k$. Hence, we wish to show that $2^2 \nmid \binom{2n}{n}$. But since $2 \nmid \frac{(n + 1)(n + 2) \cdots (2n - 1)}{1 \cdot 2 \cdots (n - 1)} = \frac{\binom{2n}{n}}{2}$, we need only show that $2^2 \nmid 2$, which is obvious. This concludes Case 1.

**Case 2: $k = 2$.**

Seeking a nice expression for $n + k$, we choose to set $n = 2^{\alpha} - 2$ for any $\alpha \in \mathbb{N}$ with $\alpha \geq 2$. Then, since $n + k = 2^{\alpha}$, we wish to show that
$$2^{\alpha} \nmid \binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}.$$
Notice that since $2n < 2^{\alpha + 1}$, the greatest power of $2$ that divides any term in the numerator or denominator of $\frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}$ is $2^{\alpha}$. Then, because the sets $\{1, 2, \cdots, n - 2\}$ and $\{n + 3, n + 4, \cdots, 2n\}$ are congruent modulo $2^{\alpha}$, we deduce that $2 \nmid \frac{(n + 3)(n + 4) \cdots (2n)}{1 \cdot 2 \cdots (n - 2)}$. Removing this fraction from the fraction $\binom{2n}{n} = \frac{(n + 1)(n + 2) \cdots (2n)}{1 \cdot 2 \cdots n}$, it suffices to show that $2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n}$. Keeping in mind that $n + 2 = 2^{\alpha}$, we see that the largest power of $2$ that divides the numerator is $2^{\alpha}$, while the largest power of $2$ that divides the denominator is $2^1$ (since $2 \mid n$). Therefore, $2^{\alpha - 1}$ is the largest power of $2$ that divides $\frac{(n + 1)(n + 2)}{(n - 1)n}$, so
$$2^{\alpha} \nmid \frac{(n + 1)(n + 2)}{(n - 1)n} \implies n + k \nmid \binom{2n}{n},$$
as desired.

Thus, the integers $k$ that satisfy the condition are all integers $k \neq 1$.

The answer is: \boxed{k \neq 1}.