= Solution
Use four-dimensional Minkowski signature $(+---)$, $\sigma^\mu=(I,\sigma^i)$ and $\bar Q_{\dot\alpha}=(Q_\alpha)^\dagger$. The odd <supercharges> transform as a <Weyl spinor> and its conjugate. In the ordinary particle <Super-Poincaré algebra>, <Lorentz covariance> makes the mixed anticommutator a vector, since $(1/2,0)\otimes(0,1/2)=(1/2,1/2)$. The even vector generator is the translation generator $P_\mu$, so the anticommutator is a real normalization constant times $\sigma^\mu P_\mu$. Unitarity makes that constant positive, and rescaling the charges sets it to two.
The same-chirality anticommutator is symmetric in its two spinor indices. A Lorentz-scalar term would require $\epsilon_{\alpha\beta}$ and hence an antisymmetric internal charge label, unavailable for $\mathcal N=1$. A term in the Lorentz generators would violate the translation Jacobi identity. Excluding additional tensorial brane charges, as in the ordinary particle algebra, the same-chirality anticommutators therefore vanish. Lorentz covariance could initially allow $[P_\mu,Q_\alpha]=a\sigma_{\mu\alpha\dot\gamma}\bar Q^{\dot\gamma}$. The <graded Jacobi identity> with $P_\mu,Q_\alpha,Q_\beta$, using their zero same-chirality anticommutator, then forces $a=0$. Thus the part of the <Super-Poincaré algebra> involving the odd generators is
$$
\boxed{\begin{gathered}
\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu,\qquad\{Q_\alpha,Q_\beta\}=\{\bar Q_{\dot\alpha},\bar Q_{\dot\beta}\}=0,\\
[P_\mu,Q_\alpha]=[P_\mu,\bar Q_{\dot\alpha}]=0.
\end{gathered}}
$$
The last identity means that a supercharge preserves momentum and mass.
The remaining odd-generator commutators express their spinor transformation under rotations and boosts. One consistent lower-index component convention is
$$
\begin{aligned}
[J_i,Q_\alpha]&=-\tfrac12(\sigma_i)_\alpha{}^\beta Q_\beta,&[K_i,Q_\alpha]&=\tfrac i2(\sigma_i)_\alpha{}^\beta Q_\beta,\\
[J_i,\bar Q_{\dot\alpha}]&=\tfrac12(\sigma_i^*)_{\dot\alpha}{}^{\dot\beta}\bar Q_{\dot\beta},&[K_i,\bar Q_{\dot\alpha}]&=\tfrac i2(\sigma_i^*)_{\dot\alpha}{}^{\dot\beta}\bar Q_{\dot\beta}.
\end{aligned}
$$
The barred formulas follow by Hermitian conjugation. Component coefficient matrices act in the dual representation; the signs therefore must be kept consistent, as in the <Jacobi sign test for supercharge components>.
One can also derive and check the odd brackets directly in <superspace>. With <left Grassmann derivatives>, take
$$
P_\mu=-i\partial_\mu,\qquad Q_\alpha=\partial_{\theta^\alpha}+i\sigma^\mu_{\alpha\dot\gamma}\bar\theta^{\dot\gamma}\partial_\mu,\qquad\bar Q_{\dot\beta}=-\partial_{\bar\theta^{\dot\beta}}-i\theta^\gamma\sigma^\mu_{\gamma\dot\beta}\partial_\mu.
$$
The two cross terms in the mixed anticommutator each give $-i\sigma^\mu\partial_\mu$, totaling $2\sigma^\mu P_\mu$. Equal-chirality terms vanish because the Grassmann derivatives anticommute and their coordinate dependence is of opposite chirality. All coefficients are independent of $x$, so the charges commute with translations. This realizes the displayed algebra rather than merely postulating its brackets.
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