I tried this approach, but I'm not sure if it would work.

Continuation of 4everwise's approach:

"Given a set $ A$ of $ n$ positive integers, consider the collection of all distinct sums..." By "all distinct sums", I think this just means that if you have 2,3,7,8 as elements of $ A$, you can't use the sum $ 10 = 2 + 8 = 3 + 7$ twice as the maximal sum of one class AND the minimal sum of the next class, even though you can find it in two different ways. Does anyone have a solution that works if some, but not all, of the $ a$'s are equal? I have not found a solution that works in this case.(It's trivial when they're all equal)[/hide]

Does this work (it is kind of along the lines of 4everwise's approach)? We proceed by induction. For a set of size 1, it is clearly true. Assume the statement holds for n=k. Now we prove it for n=k+1. Consider the set $\{a_1, a_2, \dots, a_k, a_{k+1}\}.$ Let us pick an element, WLOG, $a_{k+1}.$ The remaining elements, by the inductive hypothesis can be split into k classes with the desired condition. Now, we just place $a_{k+1}$ into the (k+1)st class. Am I missing something?

@limac: We are not to place the elements, we are to place the sums of the elements. @LawOfSigns: Yes, that is what is meant by distinct sums. So if we have say $a_1=a_2$, then we may just eliminate $a_2$ because it doesn't give us a new subset sum.

Let the elements of $A_n$ be $a_1$, $a_2$...$a_n$. We claim that letting $a_1+a_2...+a_k$ for all $1\leq k\leq n$ be the maximum element of each class, gives a construction that satisfies the problems conditions. For any subset $S_i$ of $A$, $S_i$ either belongs in one of the classes or doesn't. Suppose there exists some $k$ such that for suitable choice of $i$, we have $\sum_{p=1}^{i} a_p < S(k) < \frac{1}{2}\sum_{p=1}^{i+1} a_p$, where $S(k)$ is the sum of the elements of $S_k$. In other words, $S(k)$ is "in-between" classes. It follows that $a_1+a_2+\dots +a_i <a_{i+1}$. However this produces a contradiction. If $a_{i+1}\in S_k$ then we have that $S(k)>\frac{1}{2}\sum_{p=1}^{i+1} a_p$. If $a_{i+1}\not \in S_k$, then we see that $S(k)<\sum_{p=1}^{i} a_p < S(k)$. This also produces a contradiction, hence, we have that every subset sum belongs to a class, and we are done. $\square$

Can you Please explain with say set with 10 elements.I am having hard time understanding the class or partition of the sum into n class

Suppose $A = \{a_1 < \dots< a_n\}$ and consider partial sums $a_1$, $a_1+a_2$, $\dots$, $a_1+\cdots+a_n$. For $i = 1, \dots, n$, form the $i$th class as follows. If $a_i > a_1+\cdots+a_{i-1}$, it will be $[a_i, a_1+\cdots+a_i]$, and otherwise (if $a_i \le a_1+\cdots+a_{i-1}$) it will be $[a_1+\cdots+a_{i-1}, a_1+\cdots+a_i]$. It is easy to see that the ratio condition holds for this construction. To show that this works, let $S = \{a_{i_1} < \cdots < a_{i_{\ell}}\}$ be some non-empty subset of $A$. There is some $j \le i_{\ell}$ such that $a_1+\cdots+a_{j-1} < \sigma(S) \le a_1+\cdots+a_j$. Then $\sigma(S)$ lies in both ranges $[a_1+\cdots+a_{j-1}, a_1+\cdots+a_j]$ and $[a_j, a_1+\cdots+a_j]$ (the latter being the case since $\sigma(S) \ge a_{i_{\ell}} \ge a_j$).

Solution from Twitch Solves ISL: By induction on $n$ with $n=1$ being easy. For the inductive step, assume \[ A = \left\{ a_1 > a_2 > \dots > a_n \right\}. \]Fix any index $k$ with the property that \[ a_k > \frac{\sigma(A)}{2^k} \](which must exist since $\frac{1}{2}+\frac14+\dots+\frac{1}{2^k} < 1$). Then We make $k$ classes for the sums between $\frac{\sigma(A)}{2^k}$ and $\sigma(A)$; this handles every set which has any element in $\{a_1, \dots, a_k\}$. We make $n-k$ classes via induction hypothesis on $\{a_{k+1}, \dots, a_n\}$. This solves the problem.

Assume all sets with $n$ elements satisfy the problem conditions. We want to prove that all sets with $n+1$ elements work as well. Note that we are essentially adding an element greater than or equal to all elements in a set with $n$ elements to achieve a set with $n+1$ elements, let this number be $x$. Now consider the collection $C$ of all $\sigma(S)$ for all nonempty subsets $S$ in set $A$. Denote the least and greatest numbers in $C$ as $L$ and $G$ respectively. Also note that the new collection of the set with $n+1$ elements is $C$ + every element in $C+x$. We also need to have another class. We have the following two cases: $\textbf{Case 1: } x \ge G$ Then $2(F+x) \ge G+x$, so all elements in $C+x$ can be held in a class. $\textbf{Case 2: } x < G$ Then the maximum is $G+x$, which is also less than $2G$. Therefore, all elements within $[G,G+x]$ can be held in a class. We also know that $n=1$ works for all sets We have proved that it is possible to go from any set with $n$ elements to any set with $n+1$ elements, so we are done. $\blacksquare$

We will use induction. This is clearly true of $n=1$. Consider a set $A$ of size $n$, and suppose that this is true for all sets smaller than $n$. Let $G$ be the sum of all elements. If the largest element of $A$ is at least $\frac{1}{2}G$, we can simply resolve the $n-1$ smallest elements with $n-1$ classes by induction, and put all subsets containing the largest element in their own class since they will all be between $\frac{1}{2}G$ and $G$. Similarly, if the second largest element is at least $\frac{1}{4}G$, we can resolve the $n-2$ smallest elements with $n-2$ classes, and all subsets containing at least one of the two largest elements goes into either the $\frac{1}{4}G$ to $\frac{1}{2}G$ class or the $\frac{1}{2}G$ to $G$ class. We can repeat the same argument if the third largest is at least $\frac{1}{8}G$, by putting everything containing one of the 3 largest into one of 3 classes that collectively span from $\frac{1}{8}G$ to $G$. Therefore, if there exists any $k$ such that the $k$th largest element is at least $\frac{1}{2^k}G$, we would be done by induction. However, such a $k$ clearly exists, since if not, the sum of all elements cannot be $G$ as even if the geometric series is summed infinitely we only get $G$, hence done.

Very magical problem, if you ask me. Or maybe I just don't have combinatorics intuition... Let $a_1 < a_2 < \cdots < a_n$ be the elements of $S$. I claim that we can set the largest element of partition $A_k$ to be $p_k = a_1+a_2+\cdots+a_k$ for $1 \leq k \leq n$. Suppose there exists a sum $\sigma(A)$ that is not in one of the partitions, and suppose that $p_k < \sigma(A) < p_{k+1}$. It follows that $\sigma(A) < \frac 12 p_{k+1}$ and thus $p_{k+1} > 2p_k$, or $a_{k+1} > p_k$. But then $A$ must contain $a_{k+1}$ or a larger $a_i$, so it follows that $\sigma(A) > a_{k+1} > \frac 12 p_{k+1}$, as needed.