= Solution
Work in four-dimensional <Minkowski spacetime>, with $\bar Q_{\dot\alpha A}=(Q_\alpha^A)^\dagger$ in a <unitary representation>. The usual <Super-Poincaré algebra> has the <Poincare algebra> and possible internal <Lorentz scalars> as its even generators, and only the stated <Weyl spinors> as its odd generators. This closure assumption is essential: <Lorentz covariance> alone would also permit additional tensor-valued generators. It is the setting selected by the <Haag–Łopuszański–Sohnius theorem> for ordinary interacting relativistic theories.
The <Spinor representation of the Lorentz group> gives
$$
[Q_\alpha^A,M^{\mu\nu}]=(\sigma^{\mu\nu})_\alpha{}^\beta Q_\beta^A.
$$
Here $\sigma^{\mu\nu}$ denotes the <Lorentz algebra> matrices in the convention of the paper. Moving $M^{\mu\nu}$ to the left changes the sign; definitions using Hermitian <Lorentz algebra> generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted <Weyl spinor> indices.
The <tensor product of group representations>
$$
(\tfrac12,0)\otimes(0,\tfrac12)=(\tfrac12,\tfrac12)
$$
is a <Lorentz four-vector>. The only such even generator is $P_\mu$, so the mixed <anticommutator> must be
$$
\{Q_\alpha^A,\bar Q_{\dot\beta B}\}=H^A{}_B\,\sigma^\mu_{\alpha\dot\beta}P_\mu.
$$
Applying <Hermitian conjugation> makes $H$ a <Hermitian matrix>. Positivity of $\{S,S^\dagger\}$ for every <linear combination> $S$ of <supercharges> makes $H$ a <positive semidefinite matrix> on a positive-energy physical <superalgebra representation>. Remove any null <supercharges>, which act trivially, and use a <unitary diagonalization of a normal matrix> followed by rescaling to obtain $H^A{}_B=2\delta^A{}_B$. Thus $\mathcal N$ counts the independent nontrivial <supercharges>.
For equal <chirality>, the <tensor product of group representations> is
$$
(\tfrac12,0)\otimes(\tfrac12,0)=(0,0)\oplus(1,0).
$$
Consequently the tentative equal-<chirality> <anticommutator> can contain $\epsilon_{\alpha\beta}Z^{AB}$ and a symmetric spinor tensor $T_{\alpha\beta}^{\mu\nu}M_{\mu\nu}Y^{AB}$, where $T$ is the <intertwining operator> for <Lorentz covariance> onto $(1,0)$. Symmetry of the <anticommutator> under $(A,\alpha)\leftrightarrow(B,\beta)$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal <Lorentz scalar>, hence commutes with $P_\mu$; its full centrality will follow below.
To determine the <commutator> with translations, <Lorentz covariance> and the absence of additional odd generators leave only
$$
[P_\mu,Q_\alpha^A]=C^{AB}(\sigma_\mu)_{\alpha\dot\beta}\bar Q_B^{\dot\beta}
$$
and the relation obtained by <Hermitian conjugation>. The <graded Jacobi identity> for $(P_\mu,P_\nu,Q)$, using $[P_\mu,P_\nu]=0$, gives
$$
C\overline C\,\bigl(\sigma_\mu\bar\sigma_\nu-\sigma_\nu\bar\sigma_\mu\bigr)=0,
\qquad C\overline C=0,
$$
where $\overline C$ means entrywise <complex conjugation>. This equation alone would not imply $C=0$. Contract the <graded Jacobi identity>
$$
[P_\mu,\{Q_\alpha^A,Q_\beta^B\}]
=\{[P_\mu,Q_\alpha^A],Q_\beta^B\}+\{Q_\alpha^A,[P_\mu,Q_\beta^B]\}
$$
with $\epsilon^{\alpha\beta}$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_\mu,Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed <anticommutator> makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_\mu$. Thus $C=C^T$. It follows that $\overline C=C^\dagger$ and $CC^\dagger=0$, whence $C=0$. Applying <Hermitian conjugation> proves the same assertion for $\bar Q$:
$$
\boxed{[P_\mu,Q_\alpha^A]=[P_\mu,\bar Q_{\dot\alpha A}]=0.}
$$
Returning to the uncontracted <graded Jacobi identity> now gives $[P_\mu,\{Q_\alpha^A,Q_\beta^B\}]=0$. Since the <Lorentz algebra> generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. Hence
$$
\boxed{\{Q_\alpha^A,\bar Q_{\dot\beta B}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu\delta^A{}_B,
\qquad \{Q_\alpha^A,Q_\beta^B\}=\epsilon_{\alpha\beta}Z^{AB},
\qquad Z^{AB}=-Z^{BA}.}
$$
In particular $Z^{11}=0$ for $\mathcal N=1$. The dotted-dotted <anticommutator> follows by <Hermitian conjugation> applied to the second relation.
The <graded Jacobi identity> for $(Q^A,Q^B,\bar Q_C)$, together with $[P,Q]=0$, gives $[Z^{AB},\bar Q_C]=0$. As a <Lorentz scalar>, $Z^{AB}$ commutes with $M_{\mu\nu}$, and it already commutes with $P_\mu$. Closure and <Lorentz covariance> make $[Z^{AB},Q_\alpha^C]=D^{AB,C}{}_D Q_\alpha^D$ for some coefficients. Apply the <graded Jacobi identity> once more:
$$
0=[Z^{AB},\{Q_\alpha^C,\bar Q_{\dot\beta E}\}]
=\{[Z^{AB},Q_\alpha^C],\bar Q_{\dot\beta E}\}
=2D^{AB,C}{}_E\sigma^\mu_{\alpha\dot\beta}P_\mu.
$$
Thus $D=0$. Since every $Z$ and $Z^\dagger$ is itself an <anticommutator> of <supercharges>, the $Z$'s also commute with each other and their conjugates. Therefore \b[$Z^{AB}$ are <central charges in supersymmetry> of the displayed algebra]. Additional <R-symmetry> automorphisms are not among the generators specified here.
For <boson-fermion degeneracy in a supermultiplet>, fix a physical <four-momentum> $p$ with $p_0=E>0$, and take the finite internal space of states at that <four-momentum>. Let $\Gamma=(-1)^F$ be <fermion parity>. It anticommutes with each <supercharge> and its adjoint. For any fixed $A$, summing the diagonal spinor indices gives
$$
\sum_{\alpha=1}^2\{Q_\alpha^A,(Q_\alpha^A)^\dagger\}=4E\,\mathbf1,
$$
because $\sigma^0=\mathbf1$ and the <Pauli matrices> are traceless. Cyclicity of the <operator trace> and $\Gamma Q=-Q\Gamma$ imply
$$
\operatorname{Tr}(\Gamma QQ^\dagger)=-\operatorname{Tr}(\Gamma Q^\dagger Q).
$$
The <supertrace> of the preceding summed <anticommutator> therefore vanishes:
$$
0=4E\operatorname{Tr}\Gamma=4E(n_B-n_F),
\qquad \boxed{n_B=n_F.}
$$
This counts physical <boson> and <fermion> degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even <supersymmetric vacuum> with $P=Q=Z=0$ has one <boson> and no <fermion>. Thus the equality applies to positive-energy physical <supermultiplets> with finite state counts at fixed <four-momentum>, not to arbitrary abstract <superalgebra representations> or unregulated infinite-dimensional <operator traces>.
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