Jeff,
I do not know the chip you use. But in general, a '51 pin is like this:
There is a pull-up of the order of 40..200kOhm and a rather strong transistor to pull it down when it is in 0 state. So, when you force a 0-state pin to go up, you make it sink rather a big power that normally burns it. That is why they make it the other way: the pin is put into 1-state and the active signal applied from the exterior is ground or just low level. In this case the dissipated power is much less. It is not the question of the voltage level that you apply, but namely the polarity.
Regards,
Michael.