2005 AMC 12B Problem 20

The whole set sums to $8,$ so if one group sums to $x,$ the other sums to $8 - x$

$x^2 + (8-x)^2 = 2(x-4)^2 + 32$; check whether $x = 4$ is actually attainable

The elements sum to $8.$ If $a + b + c + d = x,$ then $e + f + g + h = 8 - x,$ so $$x^2 + (8 - x)^2 = 2(x - 4)^2 + 32.$$

This is minimized when $x = 4,$ giving $32.$ But $13$ must lie in one group, and no three of the remaining elements add with $13$ to make $4$ (that would need three of them to sum to $-9$). With $-7,$ the other two would need to sum to $-2,$ which no available pair does; without $-7,$ replacing any term in $-5-3-2=-10$ raises the sum past $-9.$ So $x = 4$ is unattainable and $(x - 4)^2 \ge 1.$

The minimum is $2(1) + 32 = 34,$ achieved for instance by $\{-7, -5, 2, 13\}$ (sum $3$) and $\{-3, -2, 4, 6\}$ (sum $5$).

Thus, the correct answer is C.