Consider the regular (n-1)-simplex $x_1+x_2+\cdots+x_n=k$ and $x_i\geq 0$. The collection of hyperplanes $x_i=p$ as well as the hyperplanes $x_i-x_j=q$ where $1\le i,j\le n$, $p.q\in \mathbb Z$, partition $\mathbb R^n$ into simplices with disjoint interiors (this is the Coxeter arrangement of type $A_{n}$, the small simplices are called alcoves). When restricted to our $(n-1)$-simplex we get a triangulation into $(n-1)$ dimensional simplices.

  These $(n-1)$ dimensional simplices also partition the positive orthant $\mathbb R_+^n$ into several cones with disjoint interior, by taking their cone from the origin. You can show that each such cone is unimodular, in the sense that its generating rays form a basis of your lattice. And this implies your claim.

Lam and Potnikov wrote in Alcoved polytopes I several equivalent combinatorial ways to get these unimodular triangulations.

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 $k$.

Consider the regular simplex $x_1+x_2+\cdots+x_n=k$ and $x_i\geq 0$. The collection of hyperplanes $x_i=j$ where $1\le i\le n$, $0\le j\le k$, partition our simplex into smaller copies of itself with disjoint interiors. This also partitions the positive orthant $\mathbb R_+^n$ into several cones with disjoint interior.

If you intersect each cone with the set of lattice points whose coordinates sum to a multiple of $k$ you can show that you get a cone isomorphic to $\mathbb N^n$. This shows that each nonnegative vector with coordinates adding to a multiple of $k$ can be written as a nonnegative combination of the $n$ generators of the cone it lies in.

Hint for the proof: The rays of the cone are given by $(a_1,a_2,\dots,a_n)+e_i$ where $1\le i\le n$. Where the $a$'s sum to $-1\pmod{k}$ and the $e_i$'s are the coordinate vectors.

Consider the regular (n-1)-simplex $x_1+x_2+\cdots+x_n=k$ and $x_i\geq 0$. The collection of hyperplanes $x_i=p$ as well as the hyperplanes $x_i-x_j=q$ where $1\le i,j\le n$, $p.q\in \mathbb Z$, partition $\mathbb R^n$ into simplices with disjoint interiors (this is the Coxeter arrangement of type $A_{n}$, the small simplices are called alcoves). When restricted to our $(n-1)$-simplex we get a triangulation into $(n-1)$ dimensional simplices.

These $(n-1)$ dimensional simplices also partition the positive orthant $\mathbb R_+^n$ into several cones with disjoint interior, by taking their cone from the origin. You can show that each such cone is unimodular, in the sense that its generating rays form a basis of your lattice. And this implies your claim.

Lam and Potnikov wrote in Alcoved polytopes I several equivalent combinatorial ways to get these unimodular triangulations.