Yes. Its π 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:
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As you can see, the first and the last resonance structures are equally as stable---both neutral, both have π bonds evenly distributed throughout the ring, and both retain the lone pair of electrons in nitrogen's perpendicular sp2 orbital.
This is important: nitrogen's perpendicular sp2 orbital contains a localized electron pair. It is not localized to and does not participate in the π orbital system, and therefore does not count towards the 4n+2 rule.
Here is a diagram showing that:
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(Had you counted those sp2 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.)
