Alice, Bob and his 1956-Triumph 
Alice challenges Bob with a puzzle and Bob accepts it even before Alice told him specifically what it is about :)
He must secretly write down a list (list A) of 20 positive rational numbers, not necessarily different from each other and put the list in a sealed envelope. Then Bob must give Alice a list (list B) of different numbers, each of which can be either one of the numbers in his list A, or the sum of more than one numbers in list A. Then Alice must try to find the numbers of list A. If she manages to find 2 or more sets of 20 numbers (from the numbers of list B), by which she can guess the numbers of list A, then Bob must donate her his priceless 1956 Triumph TR3. If, however, by the numbers in list B there is only one way to guess the 20 numbers in list A, then Bob will pay Alice one dollar for each of the numbers in list B. What is the minimum number that Bob must pay to Alice (to save his Triumph)?

This was given to me as a challenge from a friend. I am obsessed with maths and combinatorics but, alas, with this one I can't even think of where to start from! (and I'm not even sure what category to assign it to!! I chose "combinatorics" only by intuition!!)
 A: (it's essentially solution from @Vepir's comment, slightly modified to make proof simpler)
If $x \in B$, then some number from $[\frac{x}{20}, x]$ is in $A$ (we can't get $x$ by using numbers greater than $x$, or by using at most $20$ numbers smaller then $\frac{x}{20}$). So the idea is to use $20$ numbers from $B$ to make $20$ non-overlapping intervals for elements of $x$ (thus ensuring that each interval contains exactly one element from $A$), and then adding one more element to ensure all the numbers are from right border of corresponding intervals.
To do it, let $A = \{1, 10^2, 10^4, \ldots, 10^{38})$ and $B = A \cup \{1010\ldots1\}$. Let $A'$ be any list s.t. all numbers from $B$ can be sums of it. From the above argument, $A'$ contains a number from $[\frac{1}{20}; 1]$, from $[5, 100]$, from $[500, 10000]$, etc. As this intervals are non-overlapping, it contains exactly one number from each interval. If it contains any number that is not a right border of some of this intervals, then sum of even all elements of $A'$ is less than $1 + 100 + \ldots + 10^{38}$ - thus $B$ contains number that isn't sum of elements of $A'$. So $A$ contains only right borders of this intervals, so $A' = A$.
