Set p∼q when δ(p,w)∈F⟺δ(q,w)∈F for every word w. Put δ′([q],a)=[δ(q,a)] and F′={[q]:q∈F}. If p∼q, then testing continuations aw shows δ(p,a)∼δ(q,a), so δ′ is well-defined.
Take states 0,…,2n−1, transition j↦j+1(mod2n), initial state 0, and make every even state accepting. Every state is reached by some aj, and two states are equivalent exactly when they have the same parity. The quotient therefore has two states and accepts the even-lengthwords.