Equal Coins, math puzzle

An interesting problem, which has multiple solutions.

Suppose you are given $2n+1$ coins $c_1, c_2, \dots, c_{2n+1}$, of positive real weights $w_1, w_2, \dots, w_{2n+1}$.

These coins have the property that if you remove any coin, then the remaining $2n$ coins can be split into two groups of exactly $n$ coins each, such that the sum of weights of coins in one group, is same as the sum of weights of coins in the other.

Show that $w_i = w_j$ for any $i, j$.

i.e. all the coins must be of equal weight.

[Solution]