assume that you have $xRy$ and $yRz$, and then prove that you also have $xRz$.
because $xRy$ and $yRz$ then its says that $2 |(x+y)$ and also $2 |(y+z)$, then exists integers $m,k \in \Bbb Z^+$ s.t $$x+y = 2k\rightarrow x= 2k -y$$ and $$y+z=2m\rightarrow z=2m-y.$$ now, you can write that $$x+z = (2k - y) + (2m - y)= 2(k+m-y),$$ which implies that $2|(x+z)$ and therefore you have also $xRz$