An example where $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is the number of ways of counting something?
Following I. Gessel, let's denote $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ by $S(n,m)$ and call them super Catalan numbers.
In aforementioned paper Gessel proves Szily's identity: $$ S(m,n)=\sum_k(-1)^k\binom{2m}{m+k}\binom{2n}{n-k} $$ (it follows almost immediately e.g. from $(1+x)^{m+n}(1-x)^{m+n}=(1-x^2)^{m+n}$). Now, recall that without signs RHS is a Vandermonde convolution: $$ \sum_k\binom{2m}{m+k}\binom{2n}{n-k}=\binom{2(n+m)}{n+m}. $$ So $S(n,m)$ counts lattice paths from $(0,0)$ to $(n+m,n+m)$ with some signs (determined by the height of a path after first $2m$ steps).
Whether this combinatorial enough can, of course, be debated...
Introduction: The MO link provided by @JoeJohnson in the comment section clearly indicates that it is a tough problem to find a combinatorial interpretation for the positive integers \begin{align*} S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}\tag{1} \end{align*} Ira Gessel introduced for $S(m,n)$ the term Super Catalan Numbers in his paper about Super Ballot Numbers from 1992. See the answer from @GrigoryM providing some information about it. In this paper (section 6) Ira Gessel also introduces the identity
\begin{align*} \sum_{n\geq 0}2^{p-2n}\binom{p}{2n}S(m,n)=S(m,m+p)\qquad\qquad p\geq 0\tag{2} \end{align*}
which together with the initial value $S(0,0)=1$ and the symmetry $S(m,n)=S(n,m)$ shows that $S(m,n)$ is a positive integer.
Alas, he also writes that it remains to be seen whether (2) can be interpreted in a natural way.
In 2004, twelve years later(!) we can read in another paper by Ira Gessel in the introductory section:
An intriguing problem is to find a combinatorial interpretation to the super Catalan numbers.
Conclusion: A combinatorial intepretation for $S(m,n)$ is a tough problem.
Nevertheless I would like to give at least a proposal or an invitation to the reader to look for a combinatorial interpretation in terms of lattice pathes.
Observe, that $S(m,n)$ can be written as
\begin{align*} S(m,n)=\frac{\binom{2m}{m}\binom{2n}{n}}{\binom{m+n}{n}}\tag{3} \end{align*}
Interpreting this fraction in terms of lattice pathes, containing solely of $(1,0)$-steps in $x$-direction and $(0,1)$-steps in $y$-direction, we can see
the numerator \begin{align*} \binom{2m}{m}\binom{2n}{n}\tag{4} \end{align*} counts the number of pathes of length $2m+2n$ from $(0,0)$ to $(m+n,m+n)$ passing through the point $(m,m)$ or equivalently it counts the number of pathes from $(0,0)$ to $(m,m)$ times the number of pathes from $(0,0)$ to $(n,n)$.
On the other hand
the denominator \begin{align*} \binom{m+n}{m}\tag{5} \end{align*} counts the number of pathes of length $m+n$ from $(0,0)$ to $(m,n)$ or symmetrically the number of pathes from $(0,0)$ to $(n,m)$.
Idea: Since we know the fraction in (3) is a positive integer, it should be possible to find a mapping from each of the $\binom{m+n}{m}$ pathes in (5) to a set of pathes corresponding to the interpretation in (4).
For each of the $\binom{m+n}{m}$ pathes this mapping should partition the $\binom{2m}{m}\binom{2n}{n}$ pathes into equally sized classes and the challenge is to find a natural mapping for this partition.
Let's look at a simple example with $m=2,n=1$ and encode pathes using $0$ for a step in $x$-direction and $1$ for a step in $y$-direction. We observe
\begin{array}{cccc} \binom{4}{2}=6&\binom{2}{1}=2&\qquad&\binom{3}{1}=3\\ \hline 0011&01&\qquad&001\\ 0101&10&\qquad&010\\ 0110&&\qquad&100\\ 1001&&\qquad&\\ 1010&&\qquad&\\ 1100&&\qquad&\\ \end{array}
For each of the $\binom{m+n}{m}=\binom{3}{1}=3$ pathes $001,010$ and $100$ we have to find a class containing 4 pairs of pathes from $$\{0011,0101,0110,1001,1010,1100\}\times\{01,10\}$$ providing us with a natural combinatorial interpretation. At the time I wasn't able to find one, although I'm pretty sure that the information is properly encoded within these pathes.
One idea was to view the $\binom{2m}{m}$ pathes as part of a square grid in the $(x\times y)$-plane and the $\binom{m+n}{m}$ pathes within a rectangular grid of length $m$ and height $n$ as part of a $(x\times z)$-plane. This way they form two faces of a rectangular solid and the pathes are projections of an area within this solid to the faces. But I couldn't find a proper interpretation this way.
So, I invite the reader to do a small meditation exercise about the following problem:
Challenge: Imagine the $\binom{2m}{m}$ pathes are part of an $m\times m$ grid within a $(u_1\times u_2)$-plane and the $\binom{2n}{n}$ pathes are part of an $n \times n$ grid within a $(v_1 \times v_2)$-plane. Find a mapping or decipher the information encoded in the $\binom{m+n}{m}$ pathes which are part of a $m \times n$ grid in a $(w_1 \times w_2)$-plane resulting in a naturally, equally sized partitioning of the $\binom{2m}{m}\binom{2n}{n}$ pathes.