 Solution

Fermion parity is $P_f=(-1{)}^F$. An operator $O_{+}$ is parity even when $P_f{O}_{+}{P}_f=O_{+}$, equivalently $[P_f,O_{+}]=0$; it is parity odd when $P_f{O}_{-}{P}_f=-O_{-}$, equivalently $\{P_f,O_{-}\}=0$.

Odd operators supported in disjoint spacelike regions anticommute by the canonical anticommutation relations. If each were an observable, their measurement algebras would fail to commute, allowing the order of spacelike separated measurements to affect predictions. Products of an even number of fermionic fields instead commute at spacelike separation. Locality and compatible spacelike measurements therefore require every physical local observable to be fermion-parity even; parity-odd fields can create charged states but are not themselves observables.