Is the Nitrogen in pyridine electron withdrawing and what effect does this have on electrophilic substitution?
The orbitals in pyridine are pictured below; 6 pi electrons in a loop of 6 continuous p-orbitals (like benzene) plus a lone pair of electrons in an $\ce{sp^2}$ orbital on nitrogen in the plane of the ring.



Since nitrogen is more electronegative than carbon we would expect the nitrogen to inductively remove electron density from the ring carbons.  Below are the resonance structures we can draw for pyridine.  They show that the nitrogen also removes electron density from the ortho and para carbons via resonance effects. 



This removal of electron density from the ring by both inductive and resonance effects suggests that pyridine will undergo electrophilic substitution much more slowly than benzene (the ring is said to be deactivated).

If we look at the resonance structures that we can draw for electrophilic attack at the ortho



meta



and para positions,



we see that in each case we can draw 3 resonance structures, but only in the meta-attack case do we keep the positive charge on carbon and not on the more electronegative nitrogen.  Therefor, meta-substitution is preferred in electrophilic attack on pyridine



So electrophilic attack on pyridine will produce meta-substituted pyridines.  Note too that in the case of nitration, the first thing that will happen is that the lone pair on nitrogen will be protonated by the strong acid and generate the pyridinium cation.  The ring in the pyridinium cation is even more deactivated than the pyridine ring because of the full unit of positive charge on the nitrogen; but again if you look at the possible resonance structures, meta-attack will be preferred.  In this case the ring is so deactivated that extremely vigorous conditions must be used to nitrate the ring.