Let P be a ranked finite poset and suppose $x < z$ where $\textrm{rank}(z) = \textrm{rank}(x)+2$. Define the interval as $$ I_{xz} = \{y: x < y < z\} $$ Is there a simple criterion for $|I_{xz}|$ to be even? Do you have a reference? A linear order does not satisfy this property. The poset of subsets of $\{1, \ldots, n\}$ does (maybe because it is a lattice?).
Context: In case the free vector space $\mathbb{F}_2 P$ is graded, that is we have a map $\partial: \mathbb{F}_2 P \to \mathbb{F}_2 P$ with $\partial^2 = 0$, one can define its subdivision $s(P)$ as generated (over $\mathbb{F}_2$) in degree $k$ by chains of degree $k$. The differential in $s(P)$ is given by $$D(p_0 < \ldots < p_k ) = \sum_{i=0}^k p_0 < \ldots < \hat{p}_i < \ldots < p_k$$ that is, excluding one the elements of the chain. The property above (almost) ensures that the map of subdivision $$ sd(p) = \sum_{C_{\bullet} \in \max_{\le}(p) } C_{\bullet} $$ where $\max_{\le}(p)$ is the set of maximal chains with greatest element $p$, is a map of chain complexes. This is particularly interesting because the subdivision complex does not depend on the differential $\partial$ chosen on the poset.