Let $P$ be a finite poset, and let $p \in P$ be a maximal element. Then $P = (P \setminus p) \cup_{P_{<p}} P_{\leq p}$. Conversely, if $Q$ is a poset and $S \subseteq Q$ is downward-closed, then $P = Q \cup_S S^{\triangleright}$ is a poset with one more element.
Applying this iteratively, every finite poset $P$ admits an ordering $P = \{p_1,\dots, p_n\}$ such that for every $1 \leq i \leq n$, the set $\{p_1, \dots, p_i\}$ is downward-closed in $P$. In this case, $\{p_1,\dots,p_{i+1}\} = \{p_1,\dots, p_i\} \cup_{S_i} S_i^\triangleright$ for a unique downward-closed $S_i \subseteq \{p_1,\dots,p_i\}$.
Question: For which posets $P$ does there exist such an ordering where each $S_i$ is of the form $S_i = \{p_1,\dots, p_{j(i)}\}$ -- i.e. $S_i$ is downward closed not only in $P$ but also with respect to the linear ordering we've chosen?