Given a directed layered graph $G=(V,E)$, which its nodes are arranged in $n+1$ layers $L_1,...L_{n+1}$, s.t. the first and last layers have a single source and target nodes respectively ($L_1={s}, L_{n+1}={t}$). Edges are directed and only between adjacent layers, so if $e=(u,v) \in E$ then $u \in L_i$ and $v \in L_{i+1}$.
Consider the following process: We remove nodes one-by-one from the graph. If as a result of a removal, there is a different node that is no longer reachable from $s$ or that $t$ is not reachable from it, then we remove it too and so on.
The question is--what is the largest number of removal that can be applied (not including nodes that has been removed as a result of another node's removal), without disconnecting $s$ from $t$?
Is it possible to give an upper-bound for this number, depending on the number of nodes/edges/layers in the graph? (considering the worst-case order of nodes removal)
You may want to look at Menger's Theorem. This tells us that the number of nodes you can remove between $s$ and $t$ without disconnecting them is equal to the number of internally disjoint paths from $s$ to $t$ (there are directed and undirected versions of this).
Since deleting a whole layer always disconnects $s$ and $t$, there is a trivial lower bound is $\min\{|L_i|: 2\leq i \leq n\}$.
It's also easy to show that you can construct a graph with $m$ nodes, in which the smallest $s-t$ separating set has $m-2$ nodes (i.e., the whole rest of the graph, see the figure).
EDIT: It is possible to find a bound based on the order of the graph, and the number of layers. If the graph has $m$ vertices, and $n+1$ layers, then $m-2$ of the vertices (everything except $s$ and $t$) lie in the middle $n-1$ layers. Thus, by the pigeonhole principle, there is some layer $L_i$ that has at most $\lfloor \frac{m-2}{n-1} \rfloor$ vertices. Removing this whole layer disconnects $s$ and $t$, so the number of removals you can make is bounded above by $\lfloor \frac{m-2}{n-1} \rfloor$.