 Solution

Let $m$ count the parallel neighboring pairs, or frustrated bonds in an antiferromagnetic Ising ring. The periodic boundary conditions implyThere 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, givingThe 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. ThusThis 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.