The balanced criterion does not seem any less simple than finding core allocations directly, since there are an infinite number of inequalities to check. Fortunately the situation is better than that. The set of balance weights is compact, and easily seen to be convex. Thus it suffices to check only the inequalities given by the extreme weights. What are they?
The support of a weighting scheme $\lambda$ is the set of all $S\subset\nn$ such that $\lambda_S>0$. A collection $\ss$ of subsets is balanced if it is the support of a weighting scheme. The collection $\ss$ is minimal balanced if it contains no proper sub-collection which is balanced.
Lemma. A collection $\ss$ of subsets is minimal balanced iff there is only one weighting scheme that balances it.
Proof. Suppose $\lambda$ and $\delta$ are two distinct weighting schemes both with support $\ss$. Then there is some $S$ for which $\delta_S-\lambda_S<0$. Then for all $0<\alpha<1$, \begin{equation*} \alpha\delta+(1-\alpha)\lambda=\lambda+\alpha(\delta-\lambda) \end{equation*} is also a weighting scheme. For small enough $\alpha$ it is strictly positive, but for some $\alpha$ in the range one or more of the weights will be 0,which contradicts the minimality of $\ss$.