For a subset $S$ of the natural numbers $N$ and $n\in N$ let $|S\cap n|$ be the number of members of $S$ that are less than $n$. Suppose $S$ does not have upper asymptotic density $0$. That is, $$0<\lim_{m\to \infty} \sup_{n>m}\frac {|S\cap n|}{n}.$$ Suppose $(x_n)_{n\in N} $ is a decreasing sequence of positive reals such that $\sum_{n\in N}x_n=\infty.$ Is it necessary true that $\sum_{n\in S}x_n=\infty$?

I have tried to construct a counter-example, and I have also tried to prove it, and gotten nowhere at all.

This is motivated by a Q on MathExchange: If $(a_n)_n$ and $(b_n)_n$ are decreasing positive real sequences such that $\sum_nx_n$ and $\sum_nb_n$ are divergent, is it possible that $\sum_n\min (a_n,b_n)$ converges? If the answer to my Q is "yes" then the answer to that Q is "no" because at least one of $\{n:a_n\leq b_n\},\;\{n:b_n\leq a_n\}$ has upper asymptotic density of at least $1/2$.

For a subset $S$ of the natural numbers $N$ and $n\in N$ let $|S\cap n|$ be the number of members of $S$ that are less than $n$. Suppose $S$ does not have upper asymptotic density $0$. That is, $$0<\lim_{m\to \infty} \sup_{n>m}\frac {|S\cap n|}{n}.$$ Suppose $(x_n)_{n\in N} $ is a decreasing sequence of positive reals such that $\sum_{n\in N}x_n=\infty.$ Is it necessary true that $\sum_{n\in S}x_n=\infty$?

I have tried to construct a counter-example, and I have also tried to prove it, and gotten nowhere at all.

This is motivated by a Q on MathExchange: If $(a_n)_n$ and $(b_n)_n$ are decreasing positive real sequences such that $\sum_na_n$ and $\sum_nb_n$ are divergent, is it possible that $\sum_n\min (a_n,b_n)$ converges? If the answer to my Q is "yes" then the answer to that Q is "no" because at least one of $\{n:a_n\leq b_n\},\;\{n:b_n\leq a_n\}$ has upper asymptotic density of at least $1/2$.