TITLE: Positive integer combination of non-negative integer vectors
QUESTION [10 upvotes]: A vector of positive integer numbers with $n$ coordinates is given $a=(a_1,\ldots,a_n)$. It holds that $a_1+\cdots+a_n$ is divisible by some positive integer number $k$. I have checked many cases and arrived to the conjecture that one can always find at most $n$ vectors with $n$ non-negative integer coordinates such that in all the vectors the sum of the coordinates is exactly equal to $k$ and $a$ is represented as a positive integer combination of these vectors.
Example: $n=3$, $k=5$ and $a = (12,7,6)$, then the $3$ vectors satisfying above described property are $(2,2,1)$, $(5,0,0)$ and $(1,1,3)$, because $a = 3\cdot (2,2,1) + 1\cdot (5, 0, 0) + 1\cdot (1,1,3)$.
One can manually show that the conjecture holds for $k=2,3,4,5$ or $n=2$. It is also easy to see that only $n<k$ is an interesting case, $n\geq k$ can be reduced to the former.

REPLY [5 votes]: We follow Gjergji Zaimi's approach. Consider the integer points in a simplex $\Delta(n,k)$ with vertices $ke_1,\dots,ke_n$, where $e_i$ are standard basic vectors. Call a simplex formed by $n$ points $v_1,\dots,v_n\in \mathbb{Z}^n\cap \Delta(n,k)$ unimodular, if the vectors $v_i-v_j$ generate the lattice $\{(x_1,\dots,x_n):x_i\in \mathbb{Z},\sum x_i=0\}$. Note that in this case the lattice generated by $v_1,\dots,v_n$ consists of all vectors in $\mathbb{Z}^d$ with sum of coordiantes divisible by $k$.  
It suffices to prove that any point $u\in \Delta(n,k)$ (not integer in general) belongs to a unimodular simplex. Indeed, applying this to a point $\frac{k}{\sum a_i}(a_1,\dots,a_n)$ we get $n$ vectors $v_1,\dots,v_n$ such that $a$ is their non-negative linear combination and belongs to a lattice generated by them. This is what we need.
We prove it by induction in $n$. Base $n=1$, say, is clear ($n=2$ and $n=3$ are also clear, by the way, in the latter case we get a triangular lattice and a large triangle partitioned by smaller triangles.) Assume that for smaller values of $n$ this is proved. Denote $u=(u_1,\dots,u_n)=m+w=(m_1,\dots,m_n)+(w_1,\dots,w_n)$, where $m_i\in \mathbb{Z},0\leqslant w_i\leqslant 1$. Then $\sum w_i=k-\sum m_i=:d$ is non-negative integer and we may solve the problem for a point $w\in \Delta(n,d)$. We prove that $w$ lies in a unimodular simplex with vertices belonging to the set $\{0,1\}^n\cap \Delta(n,d)$ (that is, all coordinates are 0'1 or 1's), again inducting in $n$ (with obvious base). If $d>n/2$, we replace $w$ to $(1-w_1,\dots,1-w_n)$ and $d$ to $n-d$. So, we may suppose that $d\leqslant n/2$. Assume that $w_1\geqslant w_2\geqslant \dots \geqslant w_n$. Denote $p=(1,\dots,1,0,\dots,0,1)\in \{0,1\}^n\cap \Delta(n,d)$. Then $$w=a_n\cdot p+(1-a_n)\cdot \frac{w-a_n\cdot p}{1-a_n}.$$
The vector $\tilde{w}:=\frac{w-a_n\cdot p}{1-a_n}$ has $n$-th coordiante equal to 0; sum of coordinates equal to $d$.Thus if we manage to prove that all coordinates are between 0 and 1, we may do induction step (since $p$ and any unimodular simplex in $\Delta(n-1,d)$ form a unimodular simplex in $\Delta(n,d)$). Clearly all coordinates of $\tilde{w}$ are non-negative, and if they are also at most 1, unless $a_d>1-a_n$. In that case we would have $$d=a_1+\dots+a_n>(1-a_n)d+a_n(n-d)=d+a_n(n-2d)\geqslant d,$$
a contradiction.