Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different? Protons in ortho positions towards the nitro group are said to be magnetically different, even though they're chemically equivalent. Same goes for protons ortho towards the chlorine. Why is that so?
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.