Compact Surface with Genus $1$

Question

I am currently trying to solve a problem but have no idea how to go about it. The question reads:

Show that a Compact Surface with Genus $1$ has points where the Gaussian Curvature is positive, negative and points where it vanishes.

Now, by an application of Gauss Bonnet for Compact Surfaces, I got: $$\int_{S}Kd\sigma = 2\pi\cdot\chi(S)$$ Because the Genus of the Surface is given by: $$G=\frac{2-\chi(S)}{2}$$ Obviously the Euler Characteristic of this Surface is $0$. But then the Integral of the Gaussian Curvature over $S$ gives $0$. I don't know how I can conclude anything from that, or if it even means anything in the first place. If someone is willing to help me out it would be greatly appreciated.

Answer 1

EDIT

The question was wrong (with $g=0$). A round sphere gives you a counterexample. Try proving the statement for $g=1$, however. :) [You will still need to show that there is some point where the Gaussian curvature is positive. Then the rest will follow immediately.]