Do sparse DAGs can have large min-cuts? For a graph $G$, let $e(G)$ denote the number of its edges, and $c_k(G)$ the smallest number
of edges that must be removed in order to destroy all paths of length $\geq k+1$.
Note that $c_1(G)\geq c_2(G)\geq \ldots\geq c_k(G)\geq \ldots$.
Let $K_n$ be a complete graph, and $T_n$ a complete acyclic digraph (transitive tournament) on $n$
vertices; hence $e(K_n)=e(T_n)=\tbinom{n}{2}$.
 
A classical result of Erdős and Gallai
states that 
$$
c_k(K_n)=e(K_n)-\frac{kn}{2}.
$$
In contrast, for the directed acyclic analogue $T_n$ of $K_n$, we have
$$
c_k(T_n)= k\binom{n/k}{2}=\frac{e(T_n)}{k}+\frac{n}{2}\Big(1-\frac{1}{k}\Big).
$$
Proof: To show $c_k(T_n)\leq k\tbinom{n/k}{2}$, take a topological order of vertices of $T_n$: 
vertices $i$ and $j$ are adjacent iff
$i < j$. Split the vertices into $k$ consecutive intervals of length $n/k$. If we remove all
edges whose both endpoints lie in the same interval, then we destroy all paths of length $\geq k+1$. 
 Since only $k\tbinom{n/k}{2}$ edges were removed, we are done. 

 The other direction $c_k(T_n)\geq k\tbinom{n/k}{2}$ was essentially shown
by David Eppstein in this answer:
Let $C$ be a set of edges whose removal destroys all paths of length $≥k+1$ in $T_n$. Split the vertices into $t\leq k$ layers, where the $i$-th layer
contains all vertices $u$ such that the length of a longest path to $u$ in $T_n\setminus C$
 has length $i$. Since each layer is a layer of the longest-path layering,
it is independent in $T_n\setminus C$, and therefore complete in $C$.
Thus, if $n_i$ is the number of vertices in the $i$-th layer, then
the total number of edges in $C$ is at least
$\sum_{i=1}^t\tbinom{n_i}{2}\geq t\tbinom{n/t}{2}\geq k\tbinom{n/k}{2}$, as desired. 
Q.E.D. 

Motivated by this (remarkable) difference between $c_k(K_n)$ and $c_k(T_n)$, here is my 

Question:
Does $c_k(G)$ is at most "about" $e(G)/k$ for every acyclic digraph $G$?

By "about" I mean  "times some absolute constant or times some slowly growing function in $n$".

Note that the problem is only to get rid with graphs having also short paths (shorter than $k$), 
because $c_1(G)\leq e(G)/t$ holds
for any (not necessarily acyclic) digraph $G$, where $t$ is the length of a shortest source-to-target path. This is a direct consequence of
a dual to
Menger’s theorem (attributed to Robacker):
in any directed graph, the minimum length $t$ of a path is equal
to the maximum number of edge-disjoint cuts. (The proof is elementary, see e.g.
here.)

Besides being natural in itself, an affirmative answer to my question would have some interesting consequences in boolean function complexity (see this post and references herein).  
 A: I just realized that the answer
to my question is (as suspected) NO. 
Namely, in this paper, Georg Schnitger constructed a directed acyclic graph $G$ with $n$ vertices, and
$e(G)\approx n\log n$ edges such that, for every $0\leq \epsilon < 1$ and $k=n^{\epsilon}$,
we have that $c_k(G)\geq \alpha\cdot e(G)$, where $\alpha=\alpha(\epsilon)$ is a constant
depending only on $\epsilon$. This is much larger than the "desired" upper bound
$c_k(G)\leq e(G)/k$. Actually, I think that using the Kraft inequality, one can show  that
$c_k(G)=\Omega(n\log(n/k))$ holds for every $k$: show that at least $m\log m$ edges must be removed in order to disconnect any given subset of $m$ leaves, and use the argument of the proof above (haven't verified the details yet).

The graph $G$ is constructed as follows.
       (source)
Take a complete binary tree of depth $t$; hence, we have $n=2^{t+1}-1$ vertices. 
Remove all edges. Connect
each vertex with all leaves, which were previously its descendants. Direct the new
edges in the following way: the vertex receives edges from his left leaves and sends
edges to his right leaves. 

This example also shows the optimality of depth-reductions
for DAGs proved by  Erdős, Graham and Szemerédi, and generalized by
Valiant to the following important fact:

In a DAG with $m$ edges and depth (maximum length of a path) $d$, 
it is enough to take out $mr/\log d$ edges to
reduce the depth to $d/2^r$.

