Divisibility chain inside a finite-sums set ($x_i\mid x_{i+1}$)

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 $x_{i+1}\ge 2m$ 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.