Yes. Its #pi# orbital system has #p# electrons that are delocalized all throughout the ring. Also, it has #4n + 2# delocalized #p# electrons, where #n = 1#. Furthermore, it is planar due to all of its #p# orbitals being perpendicular to the ring; you can tell from how it is an analogue of benzene. Finally, it is obvious that it is cyclic. Therefore, it matches all qualifications for an aromatic compound.
If you draw all the resonance structures, you would see the following:
As you can see, the first and the last resonance structures are equally as stable---both neutral, both have #pi# bonds evenly distributed throughout the ring, and both retain the lone pair of electrons in nitrogen's perpendicular #"sp"^2# orbital.
This is important: nitrogen's perpendicular #sp^2# orbital contains a localized electron pair. It is not localized to and does not participate in the #pi# orbital system, and therefore does not count towards the #4n + 2# rule.
Here is a diagram showing that:
(Had you counted those #sp^2# electrons as #p# electrons, you would have said pyridine followed the #4n# rule where #n = 2#, which would have made it antiaromatic, but it is not.)
