= Solution
Let $W(C)=\sum_{i\in C}w_i$. If the two disjoint winning <coalitions> exchange players, the aggregate weight is conserved:
$$
W(C_1')+W(C_2')=W(C_1)+W(C_2)\ge2q.
$$
If both resulting <coalitions> lost, each weight would be strictly below $q$, making their sum strictly below $2q$. This is impossible. Thus \b[every <weighted voting game> has <swap robustness for disjoint winning coalitions>], even when the weights or quota are negative.
\b[The converse is false.] The four-player prime-cardinality game from part (a) has <swap robustness for disjoint winning coalitions>: swapping one player each way preserves each <coalition>'s <cardinality>, so in fact both winning <coalitions> remain winning. Part (a) proves it is not a <weighted voting game>.
Even imposing monotonicity would not repair the converse under the printed disjointness restriction. As a stronger counterexample, use five players and declare a <coalition> winning exactly when it contains player $1$ and contains either both $2,3$ or both $4,5$. This is a <simple cooperative game>, and all its winning <coalitions> contain player $1$, so no two are disjoint; the stated swap condition holds vacuously.
Nevertheless $\{1,2,3\}$ and $\{1,4,5\}$ win, while $\{1,2,4\}$ and $\{1,3,5\}$ lose. Both pairs contain every player the same total number of times. A weighted representation would therefore give
$$
\begin{aligned}
2w_1+w_2+w_3+w_4+w_5&\ge2q
&&\text{from the winning pair},\\
2w_1+w_2+w_3+w_4+w_5&<2q
&&\text{from the losing pair},
\end{aligned}
$$
a contradiction. This is a <trade obstruction to weighted voting> and pinpoints how weak the disjoint-only swap requirement is.
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