No one answered this week’s problem. You can read my solution below.
Let $\Sigma$ be the compact differentiable surface. Being homeomorphic to the real projective plane $\Bbb RP^2$, it has the same Euler characteristic as $\Bbb RP^2$. The projective plane has a CW-complex structure with one 0-cell, one 1-cell and one 2-cell. Hence, its Euler characteristic is $1 - 1 + 1 = 1$. By the Gauss-Bonnet theorem, the total integral of the Gaussian curvature $K$ is $\Sigma$ is $2\pi$, so by the mean value theorem, $K$ is positive at some point on $\Sigma$.