Problem: For any nonempty set $S$ of real numbers,
let $\sigma(S)$ denote the sum of the elements of $S$.
Given a set $A$ of $n$ positive integers,
consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges
over the nonempty subsets of $A$.
Prove that this collection of sums can be
partitioned into $n$ classes so that in each class,
the ratio of the largest sum to the smallest sum does not exceed $2$.

Solution: 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 $\half+\frac14+\dots+\frac{1}{2^k} < 1$).
Then
\begin{itemize}
  \ii 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 $\{a_1, \dots, a_k\}$.
  \ii We make $n-k$ classes via induction hypothesis
  on $\{a_{k+1}, \dots, a_n\}$.
\end{itemize}
This solves the problem.

\begin{remark*}
  An easy small case to motivate this is the situation $k=1$,
  which means that the largest element $a_1 \in A$
  is enormous compared to all the other elements of $A$,
  specifically $a_1 > a_2 + a_3 + \dots + a_n$.
  In that case, all the sums involving this giant $a_1$ need to be in their own class.

  The idea of the solution is to make this argument work even if $a_1$
  is not quite that enormous but the sum of the largest few numbers
  is big compared to the rest of $A$.
\end{remark*}