What causes the Inverting input of this op amp to conduct current?
With the op-amp hitting its rails, it can no longer hold its inverting output equal to its non-inverting output through feedback.
First, there's no guarantee (unless the datasheet says so) that when there's a large differential-mode voltage across the inputs that the pins won't draw lots of current.
Second, you're driving the thing from a supply in excess of 500VRMS.  I would expect that you're bringing the inverting input to the power supply rails, alternating between VCC and VEE.  In that situation then of course the chip will draw current -- in most chips, you can just assume that if you bring an input above VCC or below VEE that current will flow.
So -- use your op-amp as an op-amp, and it'll be happier.
I would suggest that since you're obviously dropping voltage here, and since the input is larger than the power supply, that you use a voltage divider followed by a much smaller gain.  I'd design things so that with the op-amp removed from the circuit none of its pins will be driven outside the range \$V_{EE} \le v \le V_{CC}\$.  That way you know that even if there's a component failure somewhere, you're much less likely to let huge voltages into your circuit.