= Solution
Keep the actual odd-$n$ <GHZ state> from the preceding calculation. Resolving its first <qubit> in the <Hadamard basis> gives
$$
|G\rangle=\frac{|+\rangle_1|\psi^-\rangle_{23}
+|-\rangle_1|\psi^+\rangle_{23}}{\sqrt2},\qquad
|\psi^\pm\rangle_{23}=\frac{|01\rangle\pm|10\rangle}{\sqrt2}.
$$
Thus the $+$ outcome leaves $|\psi^-\rangle_{23}$ and the $-$ outcome leaves $|\psi^+\rangle_{23}$, each with <probability> $1/2$. Conditional on a known result the remaining <qubits> share a maximally <entangled state>.
When the result is not revealed, the appropriate <density matrix> is the <probability>-weighted mixture from a <nonselective quantum measurement>:
$$
\begin{aligned}
\rho_{23}
&=\frac12|\psi^-\rangle\langle\psi^-|+\frac12|\psi^+\rangle\langle\psi^+|\\
&=\frac12|01\rangle\langle01|+\frac12|10\rangle\langle10|\\
&=\frac12\bigl(|0\rangle\langle0|\otimes|1\rangle\langle1|\bigr)
+\frac12\bigl(|1\rangle\langle1|\otimes|0\rangle\langle0|\bigr)\\
&=\operatorname{diag}(0,1/2,1/2,0).
\end{aligned}
$$
The cross terms cancel. The displayed convex combination of <product states> proves directly that this is a <separable quantum state>. It still has perfect <computational-basis measurement> anticorrelation, so it is not itself a <product state>. It also equals the <partial trace> $\operatorname{Tr}_1|G\rangle\langle G|$, as required by the <no-communication theorem>: a local <quantum measurement> with an unobserved outcome cannot change the remote <reduced density matrix>.
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