MAX,MAJ variants of NP complete problems
MAJ-Subset-Sum is easy to solve. A majority of partitions sum to $K$ iff $K = 0$ and all elements are equal to $0$. Indeed, let $S$ be the (multi)set, and suppose that there were some element $x \neq 0$ in $S$. For each subset $A$ of $S \setminus \{x\}$, either $\sum A \neq K$ or $\sum A + x \neq K$, and so $K$ can be the sum of at most half the subsets. We conclude that if $K$ is a sum of the majority of the subsets that $S$ consists only of zeroes, and so $K = 0$ as well.