Why does the two-electron exchange integral in Hartree-Fock theory have positive values?
The conclusion seems obvious but I just cannot find a proof.


No, I don't think it is obvious. At least I could not think about a simple algebraic proof that the exchange integral $\langle ab \vert ba \rangle$ defined as follows,
$$
\langle ab \vert ba \rangle :=
\iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}
$$
is always positive. But I could think about relatively simple physical argument that proves that at least the sum of all the exchange integrals is positive. The argument is based on the very notion of a variation principle and goes as follows.


We know that exchange integrals arise only if we use an antisymmetric product of orbitals as our trial wave function. If we use just a simple product of them there will be just Coulomb integrals in the energy expression.
Now since the true wave function is antisymmetric, if we use an antisymmetric product of orbitals as our trial wave function instead of a simple product of them, we are guaranteed to get the lower energy since we use the qualitatively right wave function.
Exchange integrals enter the energy expression with the negative sign, thus, they have to be positive, otherwise we won't get the lower energy by using an antisymmetric product of orbitals instead of a simple product.


Again, it does not prove that each and every exchange integral is positive, only that their sum is.