Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different?
The hunch lies in the definition of magnetic equivalence:


  For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same $J$ coupling to other magnetic nuclei in the molecule.


(Quoted from Hans Reich’s NMR course online syllabus; emphasis present in the original.)

What does that mean for our specific example? Compare the molecule we have in figure 1 below.


Figure 1: 1-chloro-4-nitrobenzene with labelled protons.

Needless to say as you already figured that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent (one is transformed into the other by $C_2$‡) meaning that they will display the same chemical shift; likewise for $\ce{H^2}$ and $\ce{H^4}$.

However, for magnetic equivalence we need to consider their coupling constants to each other. If $\ce{H^1}$ and $\ce{H^3}$ are magnetically equivalent, then $J_{\ce{H^1 H^2}} \overset{!}{=} J_{\ce{H^3 H^2}}$. But, $J_{\ce{H^1 H^2}}$ is an ortho coupling with a value of $J \approx 7~\mathrm{Hz}$ while $J_{\ce{H^3 H^2}}$ is a para coupling with $J \approx 0~\mathrm{Hz}$.

Thus we can easily see that $J_{\ce{H^1 H^2}} \ne J_{\ce{H^3 H^2}}$ and therefore the nuclei $\ce{H^1}$ and $\ce{H^3}$ are magnetically non-equivalent.

The same argument can be made for $\ce{H^2}$ and $\ce{H^4}$.



‡: They are also transformed into each other by the $\sigma_{\mathrm{v}1}$ plane. However, planes of symmetry create enantiotopic protons while rotations create isotopic ones. In this case — a non-chiral environment and a non-chiral solvent — enantiotopic protons cannot be distinguished and give a common signal, and thus are sometimes incorrectly called chemically equivalent. The argument is not of relevance here though, since the $C_2$ axis renders the protons isotopic thus equivalent.