 Trade obstruction to weighted voting

Suppose winning coalitions $W_1,\dots ,W_k$ and losing coalitions $L_1,\dots ,L_k$ contain each player the same total number of times:No real-weight, weak-quota weighted voting game represents those outcomes. Summing weights over the winning family gives at least $kq$, while summing over the losing family gives strictly less than $kq$. Equal player multiplicities make the two sums identical, a contradiction. This certificate works without assuming nonnegative weights.

For example, the winners $\{1,2,3\},\{1,4,5\}$ and losers $\{1,2,4\},\{1,3,5\}$ form a two-against-two obstruction. A monotone game whose minimal winning coalitions are the first two sets realizes this certificate, even though every winner contains player $1$.