Conjecture:
Given integers $0<a_1<a_2,\dots\,$ without a common divisor, such that all sums $\sum_{k=r}^q a_k\in A=\{a_1,a_2,\dots\}$. Then there is an integer $N$ such that for all integers $n>N$ it holds that $n\in A$.
I would like to see a proof or a counter-example.
I have computationally investigated some minimal sets $A$ as above which includes the primes or the squares or the exponents of two or the faculties, and it seems like $\,\mathbb Z_+\!\!\setminus\! A\,$ always is a finite set.
If $A$ is such a set that is minimal including the primes, then
$\{n\in A|n \leq 97\}=$
$\{2,3,5,7,8,10,11,12,13,15,17,18,19,20,21,23,24,25,26,28,29,30,31,32,33,35,36,$ $37,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,56,57,58,59,60,61,63,64,$ $65,66,67,68,69,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,$ $91,92,93,94,95,96,97\}$
And if $A$ is the minimal set including the faculty numbers
$1,2,6,24,120,\dots\,$ then $\{n\in A|n \leq 120\}=$
$\{1,2,3,5,6,8,9,10,11,12,14,16,17,19,20,21,22,23,24,25,26,27,28,29,30,31,32,$
$33,34,36,37,38,39,40,41,42,43,44,45,46,47,49,50,51,52,53,54,55,56,57,59,60,$
$61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,$ $86,87,88,89,90,91,93,94,95,96,97,98,99,101,102,103,104,105,106,107,108,109,$ $110,111,112,113,114,115,116,117,118,119,120\}$