Show that $\binom{2n}{n}$ is an even number, for positive integers $n$. I would appreciate if somebody could help me with the following problem

Show by a combinatorial proof that
$$\dbinom{2n}{n}$$
is an even number, where $n$ is a positive integer.

I tried to solve this problem but I can't.

Does this count?  Count the number of ways to create a subset from a set of $2n$ elements.  For each element, it's either in the set or out of the set for a total of $2^{2n}$ subsets.  Now, these subsets can have any size from $0$ to $2n$, so we have
$$\sum_{k=0}^{2n}\binom{2n}k=2^{2n}$$
Now, choosing $k$ elements to include in the subset is the same as selecting $2n-k$ elements to exclude, or
$$\binom{2n}k=\binom{2n}{2n-k}$$
Now we can simply reorder the sum and the rest is algebra
$$\sum_{k=0}^{2n}\binom{2n}k=\binom{2n}n+\sum_{k=0}^{n-1}\left[\binom{2n}k+\binom{2n}{2n-k}\right]=\binom{2n}n+2\sum_{k=0}^{n-1}\binom{2n}k=2^{2n}$$
$$\binom{2n}n=2\left[2^{2n-1}-\sum_{k=0}^{n-1}\binom{2n}k\right]$$
