For which $n$ does $\binom{2n}{n}$ divide $\binom{4n}{n}$? I found $n=1,3,4,7,10,24$ in a search up to $n=70.$ I didn't consider my software reliable enough to go much higher.
This came up while trying to find the probability of an all black card bridge hand. [If cards drawn with replacement answer would have been $1/2^{13}…$] In particular $n=13$ was not on my short list noted above. I'm wondering whether there can be a decision one way or the other about there being infinitely many $n$ that work.
 A: Claim:

If $n > 24$, then ${\large{\binom{2n}{n}}}$ does not divide ${\large{\binom{4n}{n}}}$.

Proof:

It's easily verified that the claim holds for $n < 38$.

Suppose $n\ge 38$.

Let $p\;$be the least prime such that $p > {\large{\frac{4}{3}}}n$.

Then for $n \ge 2010760$, we have $p \le  {\large{\frac{3}{2}}}n$ by Lowell Schoenfeld's generalization of Bertrand's Postulate

$\qquad$https://en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results

and for $38\le n < 2010760$, we have $p \le  {\large{\frac{3}{2}}}n$  by direct evaluation via a Maple test program.

Then from ${\large{\frac{4}{3}}}n  < p \le  {\large{\frac{3}{2}}}n$, we get
\begin{align*}
&\bullet\;p{\,\not\mid\,}n!\\[4pt]
&\bullet\;p{\,\mid\,}(2n)!\\[4pt]
&\bullet\;p^2{\,\mid\,}(3n)!\\[4pt]
&\bullet\;p^3{\,\not\mid\,}(4n)!\\[4pt]
\end{align*}

hence 
$${\large{\frac{\binom{4n}{n}}{\binom{2n}{n}}}}$$
is not an integer since identically we have
$${\large{\frac{\binom{4n}{n}}{\binom{2n}{n}}}}=\frac{n!(4n)!}{(2n)!(3n)!}$$
and for the fraction on the right, $p^3$ divides the denominator, but doesn't divide the numerator.

This completes the proof.
