Let $P$ be an integral polytope, that is, the convex hull of some points in $\mathbb{N}^d$. 


Let $p\_1,\dots,p\_m$ be *all* lattice points in $P$.


**Question:** What is the condition on $P$ that guarantees that every lattice point in the dilation $nP$ can be expressed as $k\_1p\_1 + k\_2p\_2 + \cdots + k\_n p\_n$, where the $k\_i$ are non-negative integers? Here, $n \in \mathbb{N}$
and $k\_1+k\_2+\cdots+k\_n = n$.


*Note that not all $p\_i$ need to be vertices of $P$.* 
Clearly, all *vertices* of $nP$ are expressible in this manner, since they are dilations of the vertices in $P$. 


*Remark:* The function $f(n)$ which counts lattice points in the dilation $nP$ is an (Erhart) polynomial and the $g(n)$ that counts the number of points that *can* be expressed as $k\_1p\_1 + k\_2p\_2 + \cdots + k\_n p\_n$ is *eventually* polynomial (Khovanskii).


Thus, we must impose some extra condition on the $p\_i$s to have polynomiality all the way, and also equality.


Are there some non-trivial examples of such polytopes?



Lattice points in dilated polytopes and sumsets
Your second question is known to be true by a theorem of Khovanskii: that is, $g(n)$ is 
eventually a polynomial for large $n$. Khovanskii also compares this number with the Erhart polynomial -- that is he shows that the $n$ fold sums are approximately contained in the $n$-th dilate of the polytope (this is not so hard). But I don't know of good conditions which guarantee that the Khovanskii polynomial is the same as the Erhart polynomial. For Khovanskii's work see Theorem 1 in his paper Newton polyhedron, Hilbert polynomial, and Sums of Finite sets; translation from Russian of his paper in Functional analysis & Applications 1992 (<http://www.math.jussieu.fr/~chenhuayi/conferences/okounkov/khovanskii.pdf>). 


