I'm wondering if for every set of positive integers $\{a_0, .., a_{n-1}\}\subset\mathbb{N}_+$ with $g:=\gcd\{a_i:i<n\}$ there is $M_0$ such, that for each $M\geq M_0$ there exists $\{b_i:i<n\}\subset\mathbb{N}$, satisfying $\sum_{i<n}a_ib_i=Mg$.
I'm stuck twofold. First of all, I can't even prove/disprove it for $n=2$. Secondly, easiest idea to reduce $n$ to $n-1$ would be to replace $a_0, a_1$ with something like $c\gcd\{a_0,a_1\}$, where $c$ is coprime with each $a_i,i>1$. But again, is there guaranteed triple $b_0', b_1', c$ with $a_0b_0'+a_1b_1'=c\gcd\{a_0,a_1\}$? This is instant over integers, but I can't see this over naturals. Some elementary tools like Bézout theorem, or Chinese remainder theorem specify existence of solutions, but not precise "shape" of these.
Side note: this is reformulation of more linguistic question: is there $M_L\in\mathbb{N}$ for each language $L$ over single letter (say A), such that $\{A^{M_L}\}\#\{A^g\}^*\subset L^*$ where $^*$ is Kleene star, $\#$ is concatenation, and $g=\gcd\{|w|:w\in L\}$. Previous considerations are sufficient, because it is known, that for each such $L$ there exists finite sublanguage $L'$ satisfying $L^*=L'^*$.
For yet another idea, it could be seen as subset sum problem with unlimited supply of elements. This would probably require termination proof.
First, I'll show how to prove the property for $n$ given the property for $2$ and the property for $n-1$, which reduces the problem to the $n=2$ case.
Given a subset $K$ of the natural numbers, let $\overline K$ denote its closure under addition. Note that, if $K=\{a,b\}\sqcup K'$, then $$\overline K = \overline{K'\cup \overline{\{a,b\}}}\supset \overline{K'\cup\{xa+yb\}}$$ for any integers $x,y\geq 0$. If we assume the property for all finite sets of size $2$, then $xa+yb$ can represent any sufficiently large multiple of $\gcd(a,b)$.
So, for any finite $K$ with $|K|=n$, we can choose $xa+yb=c\gcd(a,b)$ with $c$ coprime to every element of $K$. Then, by the property for $K'\cup\{c\gcd(a,b)\}$, we have that $$\overline K\supset \overline{K'\cup\{c\gcd(a,b)\}}$$ contains every sufficiently large multiple of $$\gcd(\gcd(K'),c\gcd(a,b))=\gcd(K'\cup\{a,b\})=\gcd(K),$$ as desired.
Now, to show the $n=2$ case, it suffices to (a) only prove the $\gcd(a,b)=1$ case for simplicity and (b) only show that $\overline{\{a,b\}}$ contains some number in each residue class modulo $b$, as then you can just take each class's representative and add multiples of $b$. How can you get different residue classes modulo $b$?