Finding a solution for a simple conjecture In the Linear Algebra course that I am a TA for, the professor has asked the students to be prepared to prove the following conjecture on their upcoming final:

Suppose there are $n$ students in a class with weights $W = \{w_1, w_2, \ldots, w_n\}, w_i > 0$. If any subset of $n-1$ students can be divided into two groups with equal weight, then $n$ is an odd number.

The professor provided the proof in lecture. (It makes interesting use of the properties of determinants! An outline is available here if you're interested.)
Since most of the students in the course are first years and therefore unaccustomed to formal proofs, I wanted to give them an example of $W$ that satisfied the given condition. However, the only examples I could find were trivial (i.e. $\{1, 1, 1\}$ and $\{1, 1, 1, 3, 3\}$).
My questions are:


*

*For $n=5$, is there an example of $W$ where all the $w_i$ are unique?

*For $n\in\mathbb{N}$, is it possible to enumerate all possible $W$?


There's no solution with $n=5$ and distinct weights. 
Let the weights be $a<b<c<d<e$. Removing $a$, we must have $b+c+d=e$, or $b+e=c+d$. Removing $b$, we must have $a+c+d=e$, or $a+e=c+d$. 
Each of the first pair of equations is inconsistent with each of the second pair. E.g., if you have $b+c+d=e$, then $a+e=a+b+c+d>c+d$. 
For $n=7$, you can do it with weights $1,3,5,7,9,11,13$ (exercise for the reader). 
If there's one solution, there are infinitely many, since you can just multiply each weight by any positive number you like. I'm sure that since it can be done for $n=7$, it can be done for any larger odd $n$. 
