= Solution
Let $m$ count the parallel neighboring pairs, or <frustrated bonds in an antiferromagnetic Ising ring>. The <periodic boundary conditions> imply
$$
\prod_{j=1}^N(S_jS_{j+1})=\prod_{j=1}^NS_j^2=1.
$$
There are $N-m$ antiparallel pairs, so $(-1)^{N-m}=1$: the number of antiparallel pairs must be even. For odd $N$, $m$ must consequently be odd. Each <frustrated bond in an antiferromagnetic Ising ring> raises the <energy> by $2J$ relative to a satisfied pair, giving
$$
H=-JN+2Jm.
$$
The minimum is attained at $m=1$. Choose its position among the $N$ labeled bonds, choose either orientation of one <Ising spin>, then alternate across every other bond. This constructs exactly two <spin configurations> per position. Thus
$$
\boxed{E_0=-J(N-2),\qquad g_0=2N\quad(N\text{ odd}).}
$$
This is <geometric frustration>: every local interaction prefers opposite <Ising spins>, but the closed odd ring prevents all preferences being satisfied at once. No particular location is preferred for the single <frustrated bond in an antiferromagnetic Ising ring>.
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