It is clear that $\binom{2014}{m}\binom{m}{r}=\binom{2014}{r}\binom{2014-r}{m-r}$ since both are the number of ways to choose a group of $r$ and a group of $m-r$. Then, since $ab=cd$ and $a+b=c+d$ are both only satisfied when $a=c$ and $b=d$ or $a=d$ and $b=c$. Then, returning to the original problem, we have that either $\binom{2014}{m}=\binom{2014}{r}$ or $\binom{m}{r}=\binom{2014}{r}$. In the former case, $m=r$ or $m=2014-r$. In the latter, $m=2014$. Thus, the pairs are $\boxed{(i, i) \text{ for } 1\le i\le 2014, (2014-j,j) \text{ for } 1\le j\le 1006, \text{ and } (2014,k) \text{ for } 1\le k\le 2013}$.