 Solution

Yes: this is a monotone set function. Assume $V\ne \varnothing$, as required by its definition at the empty set. If $\varnothing \ne A\subseteq B$, every weight considered in the maximum over $A$ is considered in the maximum over $B$, hence $f(A)\le f(B)$. If $A=\varnothing$ and $B\ne \varnothing$, the minimum over $V$ is at most every weight in $B$, hence again $f(A)\le f(B)$. No sign condition on the weights is needed.