= Solution
Here a supersymmetry-preserving combination means a real combination of Hermitian <Majorana spinor> <supercharges>, in a positive-norm <unitary representation> of the ordinary <Super-Poincaré algebra>. This reality qualification is essential. Let $|\psi\rangle$ be normalized and in the charge domains, and define the real <symmetric matrix>
$$
M_{ab}=\frac12\langle\psi|\{Q_a,Q_b\}|\psi\rangle.
$$
For real $u$, the Hermitian operator $q_u=u^aQ_a$ satisfies
$$
\|q_u\psi\|^2=\langle\psi|q_u^2|\psi\rangle=u^TMu.
$$
Therefore $M$ is a <positive semidefinite matrix>, and
$$
q_u\psi=0\quad\Longleftrightarrow\quad u\in\ker M.
$$
The reverse implication uses positivity: diagonalize $M$, so a zero quadratic form has zero component in every positive-eigenvalue direction. This converts the problem into <nullity of real N=1 supercharges>, not just a condition on the determinant of an operator-valued expression.
In the equivalent <Weyl spinor> form, $\{q_\alpha,q_\beta^\dagger\}=2\sigma^m_{\alpha\dot\beta}P_m$. After a fixed invertible change of real charge basis, $M$ is the realification of the Hermitian two-by-two matrix
$$
B=E I+\mathbf p\cdot\boldsymbol\sigma,\qquad
E=\langle P^0\rangle,\quad\mathbf p=\langle\mathbf P\rangle,
$$
up to the harmless spatial-sign convention. A complex <Hermitian matrix> has the same <eigenvalues> in its realification, each repeated twice. The <Pauli matrix multiplication law> gives $(\mathbf p\cdot\boldsymbol\sigma)^2=|\mathbf p|^2I$, so these <eigenvalues> are $E+|\mathbf p|$ and $E-|\mathbf p|$, each twice. Positivity forces $E\geq|\mathbf p|$. If one nonzero real $u$ annihilates the state, $M$ has a kernel: either $E=|\mathbf p|>0$, with exactly two zero <eigenvalues>, or $E=|\mathbf p|=0$, with four. Thus
$$
\boxed{\dim_{\mathbb R}\{u:(u\cdot Q)|\psi\rangle=0\}=2\ \text{or}\ 4.}
$$
For a momentum eigenstate these are the null-momentum and zero-momentum cases; the expectation-matrix proof also covers states not assumed to have definite momentum. There are no massive states preserving a real charge in this unextended algebra.
If “linear combination” instead allows arbitrary complex coefficients, the unrestricted claim is false. In a massive <fermionic Fock space> with two lowering operators $a_1,a_2$ and a spin-half <Clifford vacuum> spectator, take
$$
|\psi\rangle=\frac{|00\rangle\otimes|\uparrow\rangle+
|01\rangle\otimes|\downarrow\rangle}{\sqrt2}.
$$
Then $a_1\psi=0$. The three vectors $a_2\psi$, $a_2^\dagger\psi$ and $a_1^\dagger\psi$ are linearly independent, by their occupation numbers and spectator <spin> labels. The complex annihilator is therefore only one-dimensional. Such a lowering operator is not a Hermitian generator of preserved real <supersymmetry>. The boxed conclusion applies to the physical real-charge interpretation; it also assumes no <domain-wall charge in N=1 supersymmetry>.
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