= Solution
Make a <bipartite spin rotation>: rotate every odd site's <spin> through $\pi$ about the $x$ axis, using $U=\prod_{n\ \mathrm{odd}}e^{-i\pi S_n^x}$. This <unitary conjugation> preserves the <spin commutation relations>. At an odd site it sends $(S^x,S^y,S^z)$ to $(S^x,-S^y,-S^z)$ and exchanges the <spin raising operator> and <spin lowering operator>.
Each bond joins one rotated and one unrotated site. In the transformed operators,
$$
\mathbf S_n\cdot\mathbf S_{n+1}\longmapsto S_n^xS_{n+1}^x-S_n^yS_{n+1}^y-S_n^zS_{n+1}^z=-S_n^zS_{n+1}^z+\frac12(S_n^+S_{n+1}^++S_n^-S_{n+1}^-).
$$
Thus
$$
\boxed{H=-J\sum_n\left[S_n^zS_{n+1}^z-\frac12(S_n^+S_{n+1}^++S_n^-S_{n+1}^-)\right]}.
$$
The selected <Néel state> becomes an all-up reference state, so a single <Holstein–Primakoff transformation> convention works on both sublattices.
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