considering sums of elements of subsets of a set; proving specific property holds 
For a given set $S$ of $n$ positive integers, let $R(S)$ the set of sums of the elements of the nonempty subsets of $S$. For example, if $S = \{1, 5, 10\}$, then $R(S) = \{1, 5, 6, 10, 11, 15, 16\}$. Prove that $R(S)$ can be partitioned into $n$ subsets such that for any two terms $a < b$ in the same subset, $2a \geq b$.

Can anyone check if my existing progress is correct? If not, please tell me how this can be fixed. :) I require help on the final part of the proof, proving that there will be no "extra" cliques...

 Let the elements of $S$ be $a_1<a_2<\cdots <a_n$. Consider a graph $G$ of $2^n-1$ vertices; let each node correspond to each sum of elements of nonempty subsets. Call an edge good if the respective values assigned to the vertices follow the following rule: if $a<b$, then $2a\ge b$ or vice versa. Connect two nodes with an edge iff the two nodes are respectively good. Thus we wish to show that there exists $n$ complete graphs within $G$ (we don't care about how such complete graphs interact with each other). First order the nodes in increasing order of value; consider the following greedy algorithm: take the node corresponding to the largest number of elements in a subset (ie. the vertex that has value $a_1+a_2+\cdots + a_n$). Then from that largest vertex continue adding on vertices with smaller values until we can't (ie. we reach a vertex that has size smaller than $\left\lceil\frac{a_1+a_2+\cdots+a_n}{2}\right\rceil$). Then take those vertices away (obviously they form our first clique within $G$), and continue this algorithm until we no longer can. If, at the end of the process, we require more complete graphs, then simply split up the largest existing clique (say, of size $k$) into a clique of size $k-1$ and one of size $1$. Since $2^n-1\ge n$ for all positive $n$, we know that this will always produce "enough" complete graphs.

How can I show that there will not be "too many" complete graphs resultant from the algorithm?
Note: I used the "contest-math" tag because this problem seems like a problem that could easily appear in  a contest math setting. Thanks!
 A: We prove that the greedy algorithm works. To comment on your approach, what Calvin said is sufficient, but in more detail, studying cliques is itself a delicate subject : there are simpler things to study like independent sets and chains , and if you had constructed graphs and studied these properties, then you may have had a better chance.

To outline the greedy approach, let $a_1<a_2<...<a_n$ be the elements of $S$ and $$a_1 = b_1<...<b_N = a_1+a_2+...+a_n$$ be the set of elements of $R(S)$. The greedy approach works as follows : the first subset $S_1$ contains $b_1$ to $b_r$ , where $r = \max\{ k : b_k \leq 2b_1\}$. Then the second subset $S_2$ contains $b_{r+1}$ to $b_{t}$ where $t = \max\{k \geq r+1 : b_k \leq 2b_{r+1}\}$, and so on till the end.
We must now prove that some $S_l$, $l \leq n$ contains $b_N = a_1+a_2+...+a_n$. Then we would be done.

To do this, let us perform induction on a well chosen statement. A little bit of playing around leads to showing that $\max S_i \geq a_1+...+a_i$ for all $i = 1,2,...$. This would of course show the result.
For $i = 1$, $b_1 = a_1 \in S_1$, so of course $\max S_1 \geq a_1$.

Let's look at $i=2$. We want to say that $\max S_2 \geq a_1+a_2$. To see this, note that $a_2 \geq a_1$, so $2a_2 \geq a_1+a_2$. Note that $b_2 = a_2$,so we have $\min S_2 \geq b_2 = a_2$, and from here, since $\max S_2 \leq 2 \min S_2$ must hold true, we get $a_1+a_2 \leq \max S_2$.

Let's look at $i=3$. Again break into two cases : $a_3 \leq a_1+a_2$ and $a_3 > a_1+a_2$.
In the former case, we have $2(a_1+a_2) \geq a_1+a_2+a_3$, so $$2 \min S_3 \geq 2\max S_2 \geq 2(a_1+a_2) \geq a_1+a_2+a_3$$
therefore the greedy algorithm ensures $a_1+a_2+a_3 \leq \max S_3$.
In the latter case, we have $2a_3 \geq a_1+a_2+a_3$, and here note that if $a_1+a_2 = b_l$ then $a_3 = b_{l+1}$, because any subset sum bigger than $a_1+a_2$ will involve an $a_l$  with $l \geq 3$, hence is bigger than $a_3$. Therefore, we have $a_3 \geq \min S_3$, so from above we have $a_1+a_2+a_3 \leq \max S_3$.

Now let's do the induction. Assume that $a_1+...+a_l \leq \max S_l$. We either have $a_{l+1} \leq a_1+...+a_l$ or $a_{l+1} > a_1+...+a_l$.
In the former case, $a_1+...+a_{l+1} \leq 2(a_1+...+a_l)$ so by induction things work out.
In the latter case, if $a_1+...+a_l = b_s$ then $a_{l+1} = b_{s+1}$, for reasons similar to that mentioned in the $i=3$ case, so $a_1+...+a_{l+1} \leq 2a_{l+1} \leq 2 \min S_{l+1}$. Hence, we are done!
A: Here is a solution, although it does not extend the graph approach you have taken.
Let the $n$ integers of $S$ be $a_1 < a_2 < \cdots < a_n$, and let $s_k = \sum_{i\leq k} a_i$. Out partitioning scheme will simply be to put each $x$ into partition $k$ if $s_k/2 \leq x \leq s_k$, making the choice arbitrarily if multiple $k$ work. Such a partitioning scheme clearly will meet the required condition. It merely suffices to show that every $x\in R(S)$ will fall into at least one such partition.
To show this, assume the contrary. Suppose that for some $T \subseteq [1...n]$, that the integer $x = \sum_{i \in T} a_i$ does not fall into any such partition. Then we must have $x > s_k$ and $x < s_{k+1} / 2$ for some $k$. Substituting $s_k + a_{k+1}$ for $s_{k+1}$, we can combine these inequalities to yield $2x - a_{k+1} < s_k < x$, or $x < a_{k+1}$. But if $x < a_{k+1}$, then $T \subseteq [1\ldots k]$, and so $x \leq s_k$, a contradiction. Thus the partitioning scheme works.
As pointed out in the comments, the graph approach you present still leaves the crux of the problem to solve. It seems unlikely to me that any results of graph theory will help complete your proof. At the very least, you need to make use of the fact that the $a_i$ are positive, as the problem statement is false without this requirement. It's unclear how you can encode this requirement into the language of graphs in a meaningful way.
