Let's connect the numbers from the set $S = \{1,2,3,\ldots,2n-1,2n\}$ in pairs:

Let $X$ be any $n$-element subset of the set $S$. Choose (if possible) the smallest number $x$ from the set $X$ with the property that the number $y$ paired with $x$ does not belong to the set $X$. Then remove the element $x$ from the set $X$ and replace it with the element $y$. The resulting set is called $f(X)$. The sets $X$ and $f(X)$ have sums of elements of different parity.
In the further part of the solution, we will consider two cases:
(a) The number $n$ is odd.
The mapping $f$ is defined for any $n$-element subset of the set $S$ and it pairs all $n$-element subsets with odd sums of elements with all $n$-element subsets with even sums of elements. This proves that the number of $n$-element subsets with even sums of elements is the same as the number of $n$-element subsets with odd sums of elements.
(b) The number $n$ is even.
In this case, the mapping $f$ can only be defined for those $n$-element subsets of the set $S$ that consist of complete pairs. There are $\binom{n}{n/2}$ such subsets, and all of them have sums of elements with the same parity as the parity of the number $n/2$. The remaining subsets can be paired as $(X,f(X))$. If $n$ is divisible by four, then the number of $n$-element subsets with even sums of elements is $\binom{n}{n/2}$ more than the number of $n$-element subsets with odd sums of elements. If, however, $n$ is not divisible by four, then the number of $n$-element subsets with odd sums of elements is $\binom{n}{n/2}$ more.