What atom is always paramagnetic and why?
Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’.

The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^0$ or any other combination of two of the three equivalent p orbitals half-populated. However, in early organic chemistry courses you typically rapidly learn about the hybridisation approach which changes that electronic configuration to $\mathrm{1s^2\, 2s^1\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^1$ — four instead of two unpaired electrons. Likewise, it is possible to imagine an electronic configuration as follows: $\mathrm{1s^2\, 2s^2\,2p}_x{}^2\, \mathrm{2p}_y{}^0 2\mathrm{p}_z{}^0$. In this case, all electrons are paired and the atom is diamagnetic.

Performing this type of analysis with all the atoms at your disposal — carbon, nitrogen and oxygen — you will be able to succeed in pairing all electrons for carbon and oxygen but not for nitrogen since it has five valence electrons. Five, being an odd number, cannot be mapped pairwise. Thus, a nitrogen atom is always paramagnetic.