This question follows a previous one
If $n$ and $r$ are coprime then $a_{n,r}=\frac{1}{n}\binom{n}{r}$ is integer but this is not a necessary condition.
Question: what is a necessary and sufficient condition on $n$ and $r$ for $a_{n,r}$ to be integer?
I know that:
$$ \binom{n}{r}\equiv 0 \mod \frac{n}{\gcd {(r,n)}} $$
or in other words $a_{n,r}.\gcd {(r,n)}$ is always an integer, therefore if $n$ and $r$ are coprime then $a_{n,r}$ is is integer.
If $r=n$ or $r=0$, that is trivial. Let us have $1\le r\le n-1$. Let $p$ be a prime and $v_p(n)$ the largest exponent of $p$ such that $p^{v_p(n)}$ divides $n$
if $v_p(n)\ge v_p(r)$ then $v_p(n-r)= v_p(r)$ since $1\le r\le n-1$. Then if $r$ and $n-r$ are written in base $p$ then when adding $r$ and $n-r$ there are $v_p(n)- v_p(r)$ carries, then from Kummer's theorem $v_p(\binom{n}{r})=v_p(n)- v_p(r) $
if $v_p(n)\lt v_p(r)$ then obiously $v_p(n)- v_p(r)\lt 0 \le v_p(\binom{n}{r})$
Then whatever $n,r$ the inequality $v_p(\binom{n}{r})+v_p(r) \ge v_p(n)$ holds and the inequality $v_p(\binom{n}{r})+v_p(r) \ge v_p(r)$ also holds obviously.
Then $v_p(\binom{n}{r})+v_p(r) \ge \max(v_p(r),v_p(n))=v_p(\ lcm(r,n))$.
That is $r.\binom{n}{r}$ is multiple of $\ lcm(r,n)$.
Since $n.r= \ lcm{(r,n)}. \gcd {(r,n)}$, that is $\binom{n}{r}.\gcd {(r,n)}$ is divisible by $n$ and since $n$ is divisible by $\gcd {(r,n)}$, that is $\binom{n}{r}$ is divisible by $\frac{n}{\gcd {(r,n)}}$.
EDIT I just realized that this question is already here. The characterization of the $n,r$ such that $\gcd (n,r) \gt1$ and $ \binom{n}{r}\equiv 0 \mod {n} $ seems quite complex.
