Disjoint sets with twice ratio Given are a positive integer $n$ and positive real numbers $a_1,\dots,a_n,b_1,\dots,b_n$. A subset $S\subseteq N=\{1,\dots,n\}$ is called $a$-good if $$\sum_{i\in S}a_i\geq \frac{1}{2}\left(\sum_{i\in N\backslash S}a_i-\min_{i\in N\backslash S}a_i\right),$$
and $b$-good if $$\sum_{i\in S}b_i\geq 2\left(\sum_{i\in N\backslash S}b_i-\min_{i\in N\backslash S}b_i\right).$$
Are there always two disjoint subsets, one $a$-good and the other $b$-good?
This is true for the special case where $a_i=b_i$ for all $i\in N$: in this case, we can choose $S$ that maximizes $\min\left(2\sum_{i\in S}a_i,\sum_{i\in N\backslash S}a_i\right)$. Then $S$ will be $a$-good and $N\backslash S$ will be $b$-good. Another solution for this special case is to use a greedy algorithm, as antkam wrote in the answer below. However, neither solution seems to generalize to the original problem.

Alternate answer to the $a_i = b_i$ case / too long for a comment.
The $a_i = b_i$ case can also be solved by a greedy algorithm.  Imagine the $a_i$'s as weights, and a balance where one arm is twice as long as the other arm.  At any step: pick the heaviest remaining weight and put it on the side of the balance that is tilting up.  If the balance is even (including initially with no weights on either side), put that weight on any side.  It is easy to prove that at any time, the side tilting up would have been even or tilting down if we were to remove the latest addition to the other side (currently tilting down).  Since the latest addition is also the lightest addition (of that side), this directly proves both inequalities.
One consequence is that there must be at least $2$ solutions, since you can put the very first (i.e. globally heaviest) weight on either side of the balance to begin with.  In contrast, maximizing $\min\left(2\sum_{i\in S}a_i,\sum_{i\in N\backslash S}a_i\right)$ does not always result in multiple solutions.  E.g. if there are only two numbers $\{ \frac13, \frac23 \}$, then max-min method finds $S = \{\frac13\}$, but the greedy method would find that and also $S = \{\frac23\}$ as an alternate (and also correct) answer.  Obviously, for other examples the max-min method can find solutions that are not findable by the greedy method.  I'd say each method finds some set of solutions, and neither is a subset of another.
