In an answer to "Integer valued factorial ratios," Aaron Meyerowitz pointed out that

$$f(m,n) = \frac{(2m)! (2n)!}{m! n! (m+n)!}$$

satisfies $f(0,t) = {2t \choose t}$ and $f(i+1,j) = 4f(i,j) - f(i,j+1)$. So, by induction on the first parameter, $f(m,n)$ is an integer.

This leads to the summation

$$f(m,n) = \sum_{k=0}^m (-1)^k 4^{m-k} {m\choose k} {2(n+k)\choose n+k}.$$

This is similar, but not the same as the recurrence for $f(m,n)/2$ in Callan's paper mentioned by karan and in Yuichiro Fujiwara's answer:

$$f(m,n)/2 = \sum_{k \ge 0} 2^{n-m-2k} {n-m \choose 2k} f(m,k)/2.$$

These are both consequences of $4f(m,n) = f(m+1,n) + f(m,n+1)$ in Gessel, Super Ballot Numbers, *J. Symbolic Computation* 14 (1992), 179–194. Section 6 of that paper covers the above recurrences and more, including

$$f(m,n) = \sum_k (-1)^k {2m \choose m+k}{2n \choose n-k}$$

$$f(m,n) = (-1)^m 4^{m+n} {m-1/2 \choose m+n}.$$