If $p$ and $q$ are primes, which binomial coefficients $\binom{pq}{n}$ are divisible by $pq$? If $p$ and $q$ are primes, 
which binomial coefficients $\binom{pq}{n}$,
$1 \le n < pq$,
 are divisible by $pq$?
In particular,
if $p$ and $q$ are distinct
odd primes,
and $n$ is even,
does
$pq \mid \binom{pq}{n}$?
This question is inspired
by my answer here:
Proving that an expression returns a real non-integer number (Number 2)
According to Kummer's theorem
(https://en.wikipedia.org/wiki/Binomial_coefficient#Divisibility_properties),
"the largest power of $p$ dividing $\binom{m+n}{m}$ equals $p^c$, 
where $c$ is the number of carries when $m$ and $n$ are added in base $p$."
If $m+n = pq$,
$m = pq-n$.
So,
if there is at least one carry
when
$n$ and $pq-n$ are added
in base $p$,
then
$p \mid \binom{pq}{n}$.
So,
we want to know for which $n$
there is at least one carry
in base $p$ and base $q$.
An obvious generalization is
to ask,
if $p, q, \ldots, r$ are primes,
 which of the
binomial coefficients
$\binom{pq\cdots r}{n}$
are divisible by
$pq\cdots r$.
But I don't even know how to
do this for two primes.
Note:
The answers to this question
may be useful:
Divisibility of coefficients $\binom{n}{k}$ with fixed composite $n$
 A: For $1 \le n \le \lfloor pq/2 \rfloor$ we have

$$
\binom{pq}{n} = \frac{pq}{1} \frac{pq-1}{2} \frac{pq-2}{3} \cdots
$$

So we can divide $\binom{pq}{n}$ by $pq$
    as long as $n$ does not contain neither $p$ nor $q$,
    thus if $\gcd(pq,n) = 1$.
So at least we have

$$
\gcd(pq,n) = 1 \quad \Longrightarrow \quad pq \Bigg| \binom{pq}{n}.
$$


The general question you asked works in the same way:

$$
\gcd(p_1 p_2 p_3 \cdots p_m,n) = 1 \quad \Longrightarrow \quad
p_1 p_2 p_3 \cdots p_m \Bigg| \binom{p_1 p_2 p_3 \cdots p_m}{n}
$$

A: Observe, $\begin{pmatrix}15\\10\end{pmatrix}=3003$, which is not divisible by $15$.  This contradicts the "in particular".  Actually, the "in particular" statement is not a particular case because if $n$ is odd, then $pq-n$ is even and $\begin{pmatrix}pq\\pq-n\end{pmatrix}=\begin{pmatrix}pq\\n\end{pmatrix}$.
More generally, assume that $q>p$, then $q$ divides $(pq)!$ exactly $p$ times.  On the other hand, $q$ divides $n!$ exactly $\lfloor n/q\rfloor$ times.  Similarly, $q$ divides $(pq-n)!$ exactly $p-\lceil n/q\rceil$ times.  Since the difference between $\lfloor n/q\rfloor$ and $\lceil n/q\rceil$ is $1$, $q$ divides the binomial coefficient exactly when $n$ is not a multiple of $q$.
A similar argument will work for studying divisibility by $p$, but one must be more careful because powers of $p$ will appear.
