= Divisibility chain inside a finite-sums set
{title2=$x_i\mid x_{i+1}$}
Every <finite-sums set> generated by an infinite <sequence> of positive <integers> contains another <finite-sums set> generated by a strictly increasing <sequence> $x_i$ satisfying $x_i\mid x_{i+1}$. Consequently the divisibility requirement can be added to the conclusion of <Hindman theorem>.
Choose each $x_i$ as a sum over a finite block of indices of the original <sequence>, with these index blocks disjoint and successive. Given $x_i=m$, take two fresh groups of $m$ terms. In either group the $m+1$ <partial sums>, including $0$, give two equal <congruence classes> modulo $m$ by the <pigeonhole principle>. Their difference is a nonempty consecutive sum divisible by $m$, and hence at least $m$. Add the two such sums to obtain $x_{i+1}\ge2m$ divisible by $m$. Its index block is the <union> of the two selected subblocks. Discard the rest of these groups and continue after them. Because the blocks are disjoint, every finite sum of the $x_i$ is a finite sum of distinct original terms. This proves both the <finite-sums set> containment and the strictly increasing divisibility chain.
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