Find all $n$ for the coins 
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.

I am trying to prove this is not possible for $n = 5$, but I am not sure how to prove it. It is obvious it doesn't hold for $n=3$ since the numbers are distinct but $n=5$ seems harder.
 A: For $n=2$ it is impossible.
When $n>2$ is even it is impossible, suppose $n>2$ is even. Let $a_1,a_2\dots a_n$ be a set of number that works and let $S=a_1+a_2+\dots+a_n$, then all of the $a_i$ must have the same parity. If this parity is odd then $S-a_i$ is odd. 
Therefore each $a_i$ must be even. Now notice that if $a_1,a_2\dots a_n$ works then $\frac{a_1}{2},\frac{a_2}{2}\dots \frac{a_n}{2}$ also works. So keep dividing by $2$ until one of the elements becomes odd, then we get a set $b_1,b_2\dots b_n$ which "supposedly" works and contains an odd element. Contradiction.
So for $n$even it is impossible.
If $n=1,3$ it is clearly impossible. For $n=5$ impossible, see other post.
If $n$ is odd and at least $7$ it is possible, the set $1,3,5\dots 2k-1$ works.
A: Let the five weights be 
$$w_1 <w_2 <w_3 < w_4 <w_5$$
Now, when we remove $w_1$, we must have 
$$w_2+w_5 = w_3+w_4 \mbox{ or } \\
w_5=w_2+w_3+w_4$$
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 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 $w_2$:
$$w_5 < w_3+w_4 $$ tells us that the group with $w_5$ must contain exactly 2 coins. The second coin cannot be $w_1$, and as $w_5 >w_4 > w_1$ and the second coin is $>w_1$ we get no other choice.  
A: In fact, much more is true. This is a popular puzzle that made the rounds several decades ago. 
Theorem. If $x_1,x_2,\ldots,x_n$ are real numbers with $n$ odd, such that omitting any one of the numbers allows the others to be split into two sets of the same size and equal sum, then $x_1=x_2=\ldots=x_n$. 
Sketch proof: 
First, reduce the case of real (or even complex, doesn't matter) numbers to rational numbers. This follows from the observation that the question is really about the rank of certain rational matrices (those with 0's on the diagonal and an equal number of +1's and -1's in each row). 
Next, reduce from rationals to integers by observing that any rational solution yields an integer solution by clearing denominators. 
Finally, assuming that all the $x_i$ are integers, observe that adding a constant to all of them preserves the property, as does multiplying them all by some constant. Then, observe that in any solution, all numbers must have the same parity, which is the parity of the total sum. If they are all even, divide them all by 2; if they are all odd, add 1 and then divide by 2. Either case leads to infinite descent unless they are all equal (we may assume they are positive). 
