Supertrace pairing at positive energy ($4E(n_B-n_F)=0$)

For a finite physical supermultiplet at fixed positive energy, fermion parity anticommutes with each odd supercharge. Trace cyclicity makes $\mathrm{Tr}[(-1{)}^F\{Q,Q^{†}\}]=0$. Summing the Super-Poincaré algebra diagonal entries replaces the anticommutator by $4E$, giving equal bosonic and fermionic state counts. A zero-energy vacuum is an exception to the division by $E$, so it need not have a partner. The argument requires an invariant fixed-momentum representation, which is the step lost under explicit supersymmetry breaking.