Why am I calculating two different Norton currents for a circuit?
You are right in that the Norton (or Thevenin) impedance is the parallel combination of the two resistors.  That is 50 &Omega; in this case.

The Thevenin voltage is V1/2.  That can be seen from inspection since R1 and R2 form a voltage divider with a gain of 1/2.

You are also right that the Norton current is V1 / R1, since that's the short-circuit output current.

You should be able to derive the same Norton current by dividing the Thevenin voltage by the Thevenin (or Norton) impedance.  This also works.  That is (&frac12; V1)/(R1 // R2), which is also (&frac12; V1)/(&frac12; R1), which is V1/R1.

I didn't follow exactly what you did, but I think you might have forgotten that the Thevenin source voltage is 1/2 of V1.