Norton Equivalent and current convention
Yes, if you find that the short circuit current is negative then the short circuit current actually flows "upward" (assuming the standard way problems are drawn with two relevant nodes aligned one above the other). And if the short circuit current actually flows upward then the Norton source must be oriented pointing down, so that the current will again flow out of the lower terminal and into the upper terminal.

Maybe it is easier to think of the Thevenin equivalent situation. If you find that the open circuit voltage is negative, then you need to reverse the orientation of the Thevenin voltage source so that the open circuit voltage is negative in the Thevenin equivalent, just like it was in the original circuit.