Which half cell equation do you switch when calculating E° cell?
Now, lets first clear this up. $E_\mathrm{cell}^\circ$ can be negative, it just means that the reaction won't spontaneously occur. This means that an external current is required to be applied to push the electrons in that direction for the reaction to occur. When $E_\mathrm{cell}^\circ &gt; 0$, that just means that reaction occurs spontaneously. So in voltaic or galvanic cells, $E_\mathrm{cell}^\circ &gt; 0$.

Now to calculate $E_\mathrm{cell}^\circ$, the formula is:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} + E_\mathrm{ox}^\circ \mathrm{(anode)}$$
As you correctly stated, if you switch the direction of the reaction (so it goes from an oxidising to reducing half equation), the sign of $E^\circ$ changes. Hence the equation for $E_\mathrm{cell}^\circ$ can also be written as: 
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} - E_\mathrm{ox}^\circ \mathrm{(anode)}$$
This is the formula that you have used and hence this is why you get the correct answer when you do this.