Past exam of the computer science course of the University of Oxford / 2023 / Combinatorial optimisation / 1 / d / i / Solution

Yes, for arbitrary signed weights. A hyperedge meets both $S$ and $V\setminus S$ exactly when it meets both $V\setminus S$ and $S$. Therefore $\delta (S)=\delta (V\setminus S)$, and their weighted sums agree. The hypergraph cut function is a symmetric set function regardless of the signs of its weights.