For natural numbers $m, r$, consider the ratio of the number of subsets of size $m$ taken from a set of size $2(m+r)$ to the number of subsets of the same size taken from a set of size $m+r$:

$$R(m,r)=\frac{\binom{2(m+r)}{m}}{\binom{m+r}{m}}$$

For $r=0$ we have the central binomial coefficients, which of course are all integers:

$$R(m,0)=\binom{2m}{m}$$

For $r=1$ we have the Catalan numbers, which again are integers:

$$R(m,1)=\frac{\binom{2(m+1)}{m}}{m+1}=\frac{(2(m+1))!}{m!(m+2)!(m+1)}=\frac{(2(m+1))!}{(m+2)!(m+1)!}=C_{m+1}$$

However, for any fixed $r\ge 2$, while $R(m,r)$ seems to be mostly integral, it is not exclusively so. For example, with $m$ ranging from 0 to 20000, the number of times $R(m,r)$ is an integer for $r=2,3,4,5$ are 19583, 19485, 18566, and 18312 respectively.

I am seeking general criteria for $R(m,r)$ to be an integer.

**Edited to add:**

We can write:

$$R(m,r) = \prod_{k=1}^m{\frac{m+2r+k}{r+k}}$$

So the denominator is the product of $m$ consecutive numbers $r+1, \ldots, m+r$, while the numerator is the product of $m$ consecutive numbers $m+2r+1,\ldots,2m+2r$. So there is a gap of $r$ between the last of the numbers in the denominator and the first of the numbers in the numerator.