Let $S_n=\sum_{r=0}^n \sum_{s=0}^n \frac{(-1)^{r+s}}{{n \choose r}\left({n \choose s}+ {n \choose r}\right)}.$
Interchange $r$ and $s$, to have $S_n=\sum_{s=0}^n \sum_{r=0}^n \frac{(-1)^{s+r}}{{n \choose s}\left({n \choose r}+ {n \choose s}\right)}.$
Add these two, by symmetry, we get $$2S_n=\left(\sum_{r=0}^n \frac{(-1)^r}{n\choose r}\right)^2.$$ Finally using: Proving that $\sum_{k=0}^{n}\frac{(-1)^k}{{n\choose k}}=[1+(-1)^n] \frac{n+1}{n+2}.$
So, for even $n$, we get $$S_n=2\left(\frac{n+1}{n+2}\right)^2$$ and for odd $n$, $S_n=0,$
How else this sum can be found?
