Trivial for $(n,k)=(n,0)$ and $(n,k)=(n,n)$ So $n > k >0$ Notice that if $n$ is odd one of $k$ or $(n-k)$ is even so contradiction. And if $n$ is even then $k$ cannot be odd, because both $k$ and $n-k$ will be odd and so contradiction. Finally $n=2m$ and $k=2a$ So $\left( \binom{n}{k} \right) = \frac{n!!}{k!!(n-k)!!}= \frac{2^m m!}{2^a a!(2^{m-a} (m-a)! )}=\binom{n}{k}$

Introduce the notation $f(n,k):=\left(\binom{n}{k}\right)$. Trivially, if $k=0$ or $k=n$, then $f(n,k)$ is an integer, so from now on we assume that $0<k<n$. If $n$ is odd, then $k$ and $n-k$ have different parity, so one of them must be even. That means that the denominator of $f(n,k)=\frac{n!!}{k!!(n-k)!!} $ is even whereas the numerator is odd, so $ f(n,k)$ cannot be an integer. If $n$ is even, $n=2m$, and if $k=2j$ is also even, then using the identity $(2n)!!=2^n\cdot n!$, we find that \[f(n,k)=\frac{(2m)!!}{(2j)!!(2m-2j)!!}=\frac{m!}{j!(m-j)!}=\binom mj\in\mathbb{Z}.\] The remaining job is to check the case where $n=2m$ is even and $k$ is odd. To do this, we prove two lemmas. Lemma 1. Whenever $k<m$, we have $f(2m,k)\le f(2m,k+2)$. Proof. Using the definition, we get the following: $\frac{(2m)!!}{k!!(2m-k)!!}\le \frac{(2m)!!}{(k+2)!!(2m-k-2)!!}$. Multiplying yields $k+2=\frac{(k+2)!!}{k!!}\le \frac{(2m-k)!!}{(2m-k-2)!!}=2m-k$, which is clearly true. Lemma 2. For all $m\ge 2$ we have $f(2m,m)<2^m$ and $f(2m,m+1)<2^m$. Proof. We use induction. For $m=2$ and $m=3$, the assertion holds, as \[f(4,2)=\frac{4\cdot 2}{2\cdot 2}<2^2,\qquad f(4,3)=\frac{4\cdot 2}{3\cdot 1}<2^2,\] \[f(6,3)=\frac{6\cdot 4\cdot 2}{3\cdot 3}<2^3,\qquad f(6,4)=\frac{6\cdot 4\cdot 2}{4\cdot 2\cdot 2}<2^3.\] Suppose we are done for some $m$, then the statement for $m+2$ is obvious, because \[f(2m+4,m+2)=\frac{(2m+4)(2m+2)}{(m+2)^2}\cdot f(2m,m)<4\cdot f(2m,m)<2^{m+2},\] \[f(2m+4,m+3)=\frac{(2m+4)(2m+2)}{(m+3)(m+1)}\cdot f(2m,m+1)<4\cdot f(2m,m+1)<2^{m+2}.\] This completes the induction. Corollary. $f(2m,k)<2^m$ for any $0\le k\le 2m$ and $m\ge 2$. Proof. Since $f(n,k)=f(n,n-k)$, we may assume that $0\le k\le m$. Now perform the following algorithm - if $k<m$ then switch $k$ to $k+2$. Because of Lemma 1, $f(n,k)$ never decreases in the process. The algorithm terminates only when it reaches $k=m$ or $k=m+1$, so we find by Lemma 2 that \[f(2m,k)\le \max\{f(2m,m),f(2m,m+1)\}<2^m.\] Now suppose that for some even $n=2m$, $m\ge 2$, and odd $k$, the number $f(n,k)=\frac{2^m\cdot m!}{k!!(2m-k)!!}$ is an integer if and only if the denominator divides the numerator. Since the denominator is odd, it is relatively prime with $2^m$, hence if divides the numerator exactly if it divides $m!$. So $f(n,k)$ is an integer exactly when $\frac{f(n,k)}{2^m}$ is an integer. But the latter fraction is between $0$ and $1$, so it cannot be an integer. This proves $f(2m,k)$ also cannot be an integer for odd $k$ and $m\ge 2$. In the case $m=1$ we get one additional solution: $f(2,1)=2$ is an integer. Therefore $f(n,k)$ is an integer exactly when $0\le k\le n$ satisfy one of the following cases: $k=0$ or $k=n$, $n$ and $k$ are both even, $(n,k)=(2,1)$.