= Solution
Fix a positive-mass <unitary irreducible representation> at rest, $P^\mu=(M,0,0,0)$ with $M>0$. Use the <central charge in supersymmetry> convention
$$
\{Q^A_\alpha,\bar Q_{\dot\beta B}\}=2\delta^A_B\sigma^\mu_{\alpha\dot\beta}P_\mu,\qquad\{Q^A_\alpha,Q^B_\beta\}=\epsilon_{\alpha\beta}Z^{AB},\qquad Z^{AB}=-Z^{BA},
$$
The conjugate relation is $\{\bar Q_{\dot\alpha A},\bar Q_{\dot\beta B}\}=\epsilon_{\dot\alpha\dot\beta}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The <supercharges> commute with $P_\mu$ and the central charges. Their Lorentz transformation is that of a <Weyl spinor>; together with the ordinary <Poincare algebra>, these relations specify the <Super-Poincaré algebra>. Take $\epsilon_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2\epsilon_{\alpha\beta}Z^{AB}$ instead has $z=Z^{12}$.
When $Z=0$, define four <fermionic annihilation operators> and their <fermionic creation operators> by
$$
a^A_\alpha=\frac{Q^A_\alpha}{\sqrt{2M}},\qquad(a^A_\alpha)^\dagger=\frac{\bar Q_{\dot\alpha A}}{\sqrt{2M}}.
$$
In the rest frame they obey the <canonical anticommutation relations> $\{a_i,a_j^\dagger\}=\delta_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero <Clifford vacuum> annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the <fermionic Fock space>
$$
a_{i_1}^\dagger\cdots a_{i_k}^\dagger|\Omega\rangle,\qquad i_1<\cdots<i_k,\qquad k=0,1,2,3,4.
$$
The five levels have the <binomial coefficient> multiplicities $1,4,6,4,1$, so the <massive supermultiplet> contains $2^4=16$ states. The four creators are two copies of the spin-one-half <SU(2) representation>. Their second <exterior power> is
$$
\bigwedge^2(\mathbf2\otimes\mathbf2)=\bigl(\operatorname{Sym}^2\mathbf2\otimes\bigwedge^2\mathbf2\bigr)\oplus\bigl(\bigwedge^2\mathbf2\otimes\operatorname{Sym}^2\mathbf2\bigr).
$$
Here the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content is
$$
\boxed{5\text{ spin-0 states}+4\text{ spin-}\tfrac12\text{ doublets}+1\text{ spin-1 triplet}=16\text{ states}.}
$$
Even levels give eight <boson> states and odd levels eight <fermion> states, exhibiting <boson-fermion degeneracy in a supermultiplet>.
For nonzero $z=|z|e^{i\varphi}$, pair the two supersymmetries before normalizing. Define
$$
q_\alpha=Q^1_\alpha,\qquad r_\alpha=\epsilon_{\alpha\beta}(Q^2_\beta)^\dagger,\qquad b^\pm_\alpha=\frac{q_\alpha\pm e^{i\varphi}r_\alpha}{\sqrt2}.
$$
The rest-frame algebra gives $\{q_\alpha,r_\beta^\dagger\}=2z\delta_{\alpha\beta}$ and consequently
$$
\{b^\pm_\alpha,(b^\pm_\beta)^\dagger\}=2(M\pm|z|)\delta_{\alpha\beta},\qquad\{b^+_\alpha,(b^-_\beta)^\dagger\}=0.
$$
For $M>|z|$, divide $b^\pm$ by $\sqrt{2(M\pm|z|)}$ to recover four normalized oscillators. Positivity of the <Hilbert space> <norm> proves the <BPS bound in supersymmetry>, since
$$
\|b^-_\alpha|v\rangle\|^2+\|(b^-_\alpha)^\dagger|v\rangle\|^2=2(M-|z|)\|v\|^2\ge0.
$$
Therefore
$$
\boxed{M\ge |z|=\tfrac12|Z^{12}|.}
$$
For a <BPS state> saturating this bound, both $b^-_\alpha$ and their adjoints annihilate every state of the representation. Four of the eight real <supercharges> act trivially. Only the two $b^+$ oscillators remain, so a scalar <Clifford vacuum> gives a <shortened massive supermultiplet> with two spin-zero states and one spin-one-half doublet:
$$
\boxed{M=|z|:\quad 4\text{ states}=2\text{ bosons}+2\text{ fermions},\quad\text{one-half of the supersymmetry preserved}.}
$$
This is the irreducible count at fixed central charge. A <CPT completion of a supermultiplet> can require the charge-conjugate multiplet as well, doubling it to eight states; the full <hypermultiplet> count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.
For even $\mathcal N=2n$, <unitary skew-diagonalization of an antisymmetric matrix> brings the central-charge <matrix> to blocks $2z_r\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, $r=1,\ldots,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with <norm> $2(M+|z_r|)$ and two with <norm> $2(M-|z_r|)$, giving
$$
\boxed{M\ge\max_r|z_r|.}
$$
If $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal N-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction are
$$
\boxed{\dim\mathcal H=(2j_0+1)2^{\,2\mathcal N-2k},\qquad\frac{\text{preserved real supercharges}}{4\mathcal N}=\frac{k}{\mathcal N}.}
$$
The largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the <massless supermultiplet> construction uses a null momentum instead.
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