This is rather Combinatorics... [Amir Hossein: Thanks, moved.] If $u\le v<w$ are three such integers, we get $x=w-v,y=w-u,a=u+v-w$. So the condition can be also formulated inversely as follows: Find the biggest $g(r)$ (where $g(r)=h(r)-1$) such that there is a partition of the set $ \{1, 2,\cdots, g(r)\} $ into $r$ classes such that for all $u\le v<w$ in the same class, $u+v<w$. In particular, because of the $\le$, if $u$ belongs to a class, no element $x$ with $u<x\le 2u$ belongs to the same class. So if $g=g(r)$, the elements $g,g-1,...,[\frac{g+1}2]$ must be all in different classes. So $r\ge[\frac{g}2]+1$, which means $g(r)\le 2r-1$. Obviously, this bound is sharp, e.g. by taking sets $S_i=\{i,i+r\}$ for $i=1,...,r-1$ and $S_r=\{r\}$. Thus $h(r)=2r$.