Hmmm..... well, thanks for the responses but unfortunately I didn't understand either of them (except for the "I believe button" of adding a pull-up per Russ), but I still don't understand what's happening that would be effected by placing the pull-up.

Maybe a better question is "how is it different than the other standard I/O pins". I'm not even sure how to ask the question.

So it "drains" current I guess - from where and to where? If I put a voltate on that pin, say 3v with no resistance, will it "drain" through the PIC and out the Vss pin? Could I theoretically cook the PIC that way?

When I use a normal I/O to power an LED, it should be "draining" current from inside the PIC to that pin and down the wire to the load, so I don't know why a normal I/O wouldn't also be considered an "open drain" in this case.

Still confused.