There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that if any coin with weight $ w_k$, where $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.
My question is to prove that any $n > 5$ works. I have had difficulty trying to do this and make it precise. Here is my work to prove that $n \neq 3,5$.
Lemma 1: The number of weights $n>5$.
For $n = 3$, no such set exists since we would be left with two numbers $a,b$, which can't be equal since they are distinct.
For $n=5$, we will have $4$ numbers left which must be broken into two groups that are equal. Let the five weights be $w_1 < w_2 < w_3 < w_4 < w_5$. WLOG let's remove $w_1$. Then we must have $w_2+w_5 = w_3+w_4$ and/or $w_5 = w_2+w_3+w_4$. The second case is not possible as in this case we would have $w_5 = w_2+w_3+w_4 > w_1+w_3+w_4$, which means the coins cannot be split when we remove $w_2$. Thus we must have $w_2+w_5 = w_3+w_4$. This shows that $w_1+w_5 < w_3+w_4$. But then no splitting is possible when we remove $w_2$ since $w_5 < w_3+w_4$ tells us that the group with $w_5$ must contain exactly $2$ coins. The second coin cannot be $w_1$, and as $w_5 > w_4 > w_1$ the second coin is $> w_1$ we have no other choice.
Here is my proof that any odd $n$, $n>5$ works:
Lemma 2: Any odd $n > 5$ works.
Let $f = 1*3*\cdots*(2n-1)$ for some odd number $n$ and the $*$ symbol can take either $+,-$ where the weights of the coins are $1,3,\ldots,2n-1$. Now suppose that we can satisfy the condition of the problem for some $2n+1$ and then we need to show that $2n+3$ is possible. Then define a transformation on $f$ call it $f'$ where $\left |f' - f\right| = 2$ which we form by taking two odd consecutive integers and changing their sign to form $f'$. We then move a term from the set $\{1,\ldots,2n-1\}$ then by using the result $\left |f'-f\right|$ we change the sign on $2n+1$ or $2n+3$ using $\left|f'-f\right| = 2$ to get $f = 0$. Thus two sets have equal sum if we move $2n+3$ or $2n+1$ which holds for $n \equiv 1 \mod 4$ or $n \equiv 3 \mod 4$.
Comments
As you can probably tell, my first lemma is a bit lengthy, and so I would like to shorten it. In my second lemma it is very imprecise right now, but I think I have the right idea. Can anyone help me make that more precise? Also, do the weights need to be positive or can they be $0$?