For any two integers $[[(n_1,n_2)]]$ and $[[(n_3,n_4)]]$, strict ordering is defined as $[[(n_1,n_2)]]<_{\mathbb{Z}}[[(n_3,n_4)]]{\iff}n_1+_{\mathbb{N}}n_4<_{\mathbb{N}}n_2+_{\mathbb{N}}n_3$. Considering that $x\,{\preceq}\,y$ is often defined as $(x\,{\prec}\,y)\,{\lor}\,(x=y)$, I believe that we can extend the definition to $[[(n_1,n_2)]]\,{\preceq}_{\mathbb{Z}}\,[[(n_3,n_4)]]{\iff}n_1+_{\mathbb{N}}n_4\,{\preceq}\,_{\mathbb{N}}n_2+_{\mathbb{N}}n_3$.
It can be easily proved that the latter is antisymmetric and reflexive, but I have problem with checking transitivity. How to do it?
Assume $[[(n_1,n_2)]]\preceq_{\mathbb{Z}}[[(n_3,n_4)]]\preceq_{\mathbb{Z}}[[(n_5,n_6)]]$. That means that $n_1+n_4\leq n_2+n_3$ and $n_3+n_6\leq n_4+n_5$. Then $$n_1+n_4+n_6\leq n_2+n_3+n_6\leq n_2+n_4+n_5$$ So $n_1+n_6\leq n_2+n_5$, as required.
(I used the fact that $a+b\leq a+c\Rightarrow b\leq c$, which can be easily proved).