 Solution

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.