= Solution
For global four-dimensional $\mathcal N=1$ <supersymmetry>, use
$$
\{Q_\alpha,Q_\beta^\dagger\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu,\qquad
\{Q_\alpha,Q_\beta\}=0,\qquad [P_\mu,Q_\alpha]=0.
$$
Let $\Pi=(-1)^F$ 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> gives
$$
\operatorname{Tr}(\Pi Q_\alpha Q_\alpha^\dagger)
=\operatorname{Tr}(Q_\alpha^\dagger\Pi Q_\alpha)
=-\operatorname{Tr}(\Pi Q_\alpha^\dagger Q_\alpha).
$$
Thus the <trace> with $\Pi$ of every displayed diagonal <anticommutator> vanishes. Summing the two components, $\sum_\alpha\{Q_\alpha,Q_\alpha^\dagger\}=4E\mathbf1$, so
$$
0=4E\operatorname{Tr}\Pi=4E(N_B-N_F).
$$
Since $E>0$, this proves \b[$N_B=N_F$], 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>.
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