= Solution
Each bond of the classical <Heisenberg antiferromagnet> is minimized when its two <spins> are antiparallel. Since the periodic chain has an <even> number of sites, all bonds can be minimized simultaneously:
$$
\mathbf S_n=(-1)^nS\widehat{\mathbf n},\qquad |\widehat{\mathbf n}|=1,\qquad \boxed{E_{\mathrm{cl}}=-NJS^2}.
$$
The <ground-state degeneracy> is a continuous sphere of orientations $\widehat{\mathbf n}\in S^2$. Choose the <Néel state> with even sites in $|S,S\rangle$ and odd sites in $|S,-S\rangle$.
This product state is not an <energy eigenstate>. The <spin ladder operators> in
$$
\mathbf S_n\cdot\mathbf S_{n+1}=S_n^zS_{n+1}^z+\frac12(S_n^+S_{n+1}^-+S_n^-S_{n+1}^+)
$$
produce states with both neighboring magnetic quantum numbers changed. On an up-down bond, $S_n^-S_{n+1}^+$ has a nonzero matrix element $2S$, hence coefficient $JS$ in the <Hamiltonian>. These orthogonal spin configurations show explicitly why the classical minimum is not an exact quantum eigenstate for $S>0$.
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