Prove a coin game is fair 
Suppose $2n$ identical, equal valued fair coins are flipped. If no more than half land head, you win all the heads. If more than half land head, you lose all the heads. Prove the game is fair with combinatorial argument only.

This is derived from a recently closed question, which was linked to Prove that $\sum\limits_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}$.
If the game is fair then
$$\sum_{i=0}^{n} i \binom{2n}{i} = \sum_{j=n+1}^{2n} j \binom{2n}{j}\\ 
\implies n\binom{2n}{n} + \sum_{i=0}^{n-1} i \binom{2n}{i} = \sum_{i=0}^{n-1} (2n-i) \binom{2n}{2n-i}\\
\implies n\binom{2n}{n} = \sum_{i=0}^{n-1} 2(n-i)\binom{2n}{i} = \sum_{i=0}^{n-1} 2i \binom{2n}{n-i}$$
I found that it reduced to proving this simple symmetry:
$$(n-i) \binom{2n}{n-i} = (n+i+1) \binom{2n}{n+i+1}, \forall 0 \le i \le n-1 \tag 1$$
i.e., your gain with $2n$ heads is offset by your loss with $2n+1$ heads, gain with $2n-1$ heads is offset by loss with $2n+2$ heads, and so on, and when all land tail you breakeven.
Algebraically $(1)$ is trivially true but is there a pure combinatorial argument? Or better yet, an argument to prove the game is fair without breaking it down?
 A: Are you familiar with the combinatorial proof that
$$
k\binom{n}k=n\binom{n-1}{k-1}?
$$
Both sides answer the question: "How many ways are there to make a committee of $k$ people from a pool of $n$ people, and then make one of the committee members president?"
Applying this to each of the equation $(n-i) \binom{2n}{n-i} = (n+i+1) \binom{2n}{n+i+1}$, you get
$$
2n\binom{2n-1}{n-i-1}=2n\binom{2n-1}{n+i},
$$
which is obviously true by symmetry.
A: Based on Mike Earnest's answer I have an argument to prove the following:
$$\sum_{i=0}^{n} i \binom{2n}{i} = \sum_{j=n+1}^{2n} j \binom{2n}{j}$$
Now the question is: if there are no more than half heads, paint one of them red; otherwise paint one of more than half heads green. Show that there are same number of ways of getting a red head vs. a green head.
Now we pick a particular coin. We look at each case where it is the red coin. That means there are $k$ other heads and $2n-1-k$ tails where $0 \le k \le 2n-1$. If we flip all other $2n-1$ coins then we have $2n-k$ heads where $n+1 \le 2n-k\le 2n$ which corresponds to a case where this particular coin is a green head. And this map is one-to-one, therefore we have the same number of cases where this particular coin is a red head vs. a green head. And this is true for each and every one of the $2n$ coins. We are done.
