= Solution
At fixed nonzero <four-momentum>, take the finite physical <supermultiplet> and write $P_F=(-1)^F$ for <fermion parity>. Each <supercharge> is odd, so $P_FQ_\alpha=-Q_\alpha P_F$ and the same holds for its adjoint. By the <cyclic property of the trace>,
$$
\operatorname{Tr}(P_FQ_\alpha Q_\alpha^\dagger)
=\operatorname{Tr}(Q_\alpha^\dagger P_FQ_\alpha)
=-\operatorname{Tr}(P_FQ_\alpha^\dagger Q_\alpha).
$$
Consequently $\operatorname{Tr}[P_F\{Q_\alpha,Q_\alpha^\dagger\}]=0$. Sum the two spinor components of the $\mathcal N=1$ <Super-Poincaré algebra>. Since $\operatorname{Tr}_{2\times2}\sigma^i=0$, their sum is $4H$. On a fixed-energy representation, $H=EI$, yielding
$$
0=4E\operatorname{Tr}P_F=4E(n_B-n_F).
$$
For a nonzero physical one-particle momentum, $E>0$, including the massless case. Hence \b[the bosonic and fermionic physical state counts agree]:
$$
\boxed{n_B=n_F.}
$$
Equivalently a nonzero normalized <supercharge> and its adjoint define an odd <involution> pairing the two subspaces. This is <supertrace pairing at positive energy>; it counts spin polarizations, not merely distinct names of particles.
The phrase “every supermultiplet” requires this positive-energy qualification. A zero-energy <supersymmetric vacuum> with all <supercharges> acting as zero is a one-dimensional bosonic representation with no fermionic partner. It is a counterexample to the unrestricted statement, because division by $E$ is then impossible. It does not contradict the requested equality for particle <supermultiplets>.
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