Positive integer combination of non-negative integer vectors

Consider the regular (n-1)-simplex $x_1+x_2+\cdots+x_n=k$ and $x_i\geq 0$. The collection of hyperplanes $x_i=p$ where $1\le i\le n$, $p\in \mathbb Z$, partition our simplex into smaller polytopes with disjoint interiors. These polytopes are alcoved polytopes in the sense of Lam and Postnikov, and therefore have unimodular triangulations. By taking the cones over these triangulations you get a unimodular decomposition of $\mathbb N^n$ intersected with your lattice of vectors whose coordinates add to a multiple of $k$.

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.