I need to see for which $S$ connected, orientable (and compact) surfaces $\iint_{S} |K| dA=4\pi$, where $K$ means Gaussian curvature.
I think that it is only true if $S$ is a sphere. I have done the following:
We define $$S^{+}=\{p \in S: K(p)\geq 0\},S^{-}=\{p \in S: K(p)\leq 0\}. $$ So $$ \iint_S |K| dA= \iint_{S^{+}} K dA+ \iint_{S^{-}} (-K) dA, $$ which implies $$ \iint_{S^{+}} K dA= \iint_S |K| dA\ \underbrace{+\iint_{S^{-}} K dA}_{\leq 0}\leq \iint_S |K| dA. $$
Apart from that, $$\iint_S |K| dA = A(\mathbb{n}(S))\geq A(\mathbb{S^2})=4\pi $$
**Note**:A() means surface and n normal vector.
So $\iint_{S^{+}} K dA \geq 4\pi. $ Because of that, if we suppose that $$ \iint_S K dA = 4\pi \implies 4\pi = \iint_S |K| dA = \iint_{S^{+}} K dA $$
So we have that $K(p)\geq 0$, $\forall p \in S$.
I want to see that $K$ is, in fact, constant but I don't know how to prove it... I would really thank some help.
For surfaces with everywhere positive curvature, an easy answer comes from the Gauss-Bonnet theorem (https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem) which says that this integral is $2\pi \chi(S)$, so that the Euler characteristic, $\chi$, must be $2$. The only compact orientable surface with Euler characteristic $2$ is the sphere.
Gauss-Bonnet also tells you that for any smooth embedding of a sphere in 3-space, you get the integrated curvature is $4\pi$, so it's true even for a non-spherical ellipsoid, so your conjecture that $K$ must be constant is false.
==== For surfaces with both positive and negative curvature parts, you know that $$ \iint K \le \iint |K| $$ which says that no surface with Euler characteristic greater than 2 can have the desired property...but since there are no compact orientable surface with $\chi > 2$, that doesn't tell us much.
So I guess that this answer doesn't really get you where you want to go, alas.