Hyperbolic paraboloid $S$ is given by $$ f(x,y)=(x,y,xy)$$
Consider $$ (x,0,0),\ (0,y,0)$$ which are geodesics in $\mathbb{R}$
Hence they are geodesics on $S$
In further $$f_x=(1,0,y),\ f_y=(0,1,x),\ E=1+y^2,\ G=1+x^2,\ F=xy
$$
\begin{align*} dA&=\sqrt{1+r^2}dxdy\\
\int_S K&= \int \frac{-1}{(1+r^2)^2} \sqrt{1+r^2} dxdy \\&=\int
\frac{-1}{(1+r^2)^\frac{3}{2}}\ rdrd\theta =2\pi
(1+r^2)^\frac{-1}{2}|_0^\infty=-2\pi\ \ast\end{align*}
Consider a geodesic triangle $\Delta$ : $(t,0,0),\ (0,t,0),\
(0,0,0),\ t>0$
Then $\int_\Delta K +\sum_i \theta_i=2\pi$ so that $$
\frac{\pi}{2}<\int_\Delta K+ \pi=\sum_i(\pi-\theta_i)$$
Here first inequality is followed since we have $\int_\Delta K>
\frac{1}{4}\int_S K$ by observing integrand in $\ast$ And last term
is sum of internal angles of $\Delta$ So we can not have a
contradiction