Let $A$ be a matrix with entries either 0 or 1, where each column contains at least one 1, to remove trivial degenerations.


Let $P$ be the *convex hull of all integer vectors* $x$ that satisfy $Ax \leq y$, and $x\geq 0$, where $y$ is some non-negative integer vector. Clearly, $P$ is an integral polytope.


For example (to address David Speyers comment), when 
$$A=\begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}, y=(1,1,1)$$
then $P$ is the convex hull of the solutions to $Ax\leq y$, so $P$ is the convex hull of $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$, the standard simplex. 


Doing some computer experiments, I believe the following:


*Conjecture:* P is integrally closed, i.e., every *integer* point $p \in kP$
can be expressed as $p=p\_1+p\_2+\dots+p\_k$ where all $p\_i$ are *integer* points in $P$,
whenever $k$ is a natural number.


In the example above, this is known to be integrally closed.


Note that there are no conditions on the minors of $A$.


Is this a known result? This seems hard, since we do not have a nice description of $P$, that is, the supporting hyperplanes, nor the vertices, are explicitly known.



Integrally closed polytopes from 01-matrices
No.
$$x\_1+x\_2 \leq 1 \quad y\_1 + y\_2 \leq 1 \quad z\_1 + z\_2 \leq 1$$
$$x\_1+y\_1+z\_1 \leq 2 \quad x\_2+y\_2+z\_1 \leq 2 \quad x\_2+y\_1+z\_2 \leq 2 \quad x\_1 + y\_2 + z\_2 \leq 2$$
$$(1,1,1,1,1,1) \in 2 P.$$


Note that the first three inequalities imply $x\_1+x\_2+y\_1+y\_2+z\_1+z\_2 \leq 3$. 
If we are to sum up two such points and get a point whose coordinates sum to $6$, then the two points of $P$ must have 
$$x\_1+x\_2+y\_1+y\_2+z\_1+z\_2 = 3.$$
So we will add this equality to our list of relations. 


Together with this linear inequality, the first line of inequalities cuts out a cube. Projecting onto the $(x\_1, y\_1, z\_1)$ coordinates, it is $[0,1]^3$.


The second line of inequalities gives a tetrahedron whose vertices are four non-adjacent vertices of the cube. Projecting onto $(x\_1, y\_1, z\_1)$ again, we are talking about $(0,0,0)$, $(1,1,0)$, $(1,0,1)$ and $(0,1,1)$.


Since $x\_1+y\_1+z\_1$ is even for all integer points in $P$, it is impossible for $(1,1,1,1,1,1)$ to be the sum of two such points.


The tetrahedron formed by nonadjacent vertices of a cube is the standard example of a non-integrally closed polytope with integer vertices; I just had to figure out how to embed it in inequalities of the form you gave.


