I cannot believe the following infimum unequals zero. Can anybody find a way to construct a sequence of subsets for which the expression in the infimum tends to zero?
$$\inf\left\{\frac{\sup_{B\subseteq A}(|B|\cdot\min B)}{\sum_{a\in A} a}:A\subseteq\mathbb{R}^+\text{ countable and }\sum_{a\in A} a<\infty\right\}$$
where $|B|$ is the number of elements in $B$.
Set $A_n=\{\frac{n}{n},\frac{n}{n-1},\ldots,\frac{n}{1}\}$. That means for any $B \subseteq A_n$ with $\min B=\frac{n}i$ we have $\lvert B\rvert \le i$ and hence $\lvert B\rvert\min B \le n$ with equality achievable ($B=A_n$).
So we have
$$\frac{\sup_{B \subseteq A_n}(\lvert B\rvert\min B)}{\sum_{a\in A_n}a} = \frac{n}{n(\frac1n+\frac1{n-1}+\ldots + \frac11)}=\frac1{\frac11+\frac12+\ldots+\frac1n}.$$
This is the inverse of the partial sum of the harmonic series, so tends to $0$ for $n \to \infty$. So $A_n$ is a sequence of sets that shows the infimum is indeed $0$.