Let $(G,\cdot)$ be a group of order $2n$ which has exactly $n$ elements of order $2$. Prove that $n$ is odd and that $G$ has an abelian subgroup of order $n$.
For the first part, the elements whose order is greater than $2$ can be grouped in pairs of the form $\{x,x^{-1}\}$. Now it easily follows that $n=\text{even}-\text{odd}=\text{odd}$.
For the second part, I though about considering the set $H=\{x_1,x_2,\dots ,x_n\}$ where $x_i$ is an element of order $2$ $\forall i=\overline{1,n}$.
I observed that for $i\neq j$ we have that $x_i x_j \in G\setminus H$, because otherwise $\{e,x_i,x_j, x_i x_j\}$ would be a subgroup of order $4$ of $G$, which would contradict Lagrange's theorem. I didn't know how to proceed from here.
Best Answer
I will try to keep to the spirit of your approach. Let's write $K$ for your $G \setminus H$, the set of elements of order other than $2$. It's a good bet that $K$ should be such a subgroup, since the whole question makes it sound like your group behaves sort of like the dihedral group of order $2n$, where the $n$ reflections have order $2$ and the $n$ rotations are a cyclic subgroup.
Note that $x_1 x_1, x_2 x_1, \dotsc, x_n x_1$ are all distinct, since $x_i x_1 = x_j x_1 \implies x_i = x_j$. Therefore each element of $K$ can be written as $x_i x_1$ for some $i$. Then we can use the efficient subgroup criterion: if $a = x_i x_1$, $b = x_j x_1$ are two arbitrary elements of $K$, then $ab^{-1} = x_i x_1 x_1 x_j = x_i x_j \in K$, and also $e \in K$, so $K$ is a subgroup.
Then by this question, we're done. I've copied over the important bit for completeness:
Let $a, b \in K$ be arbitrary and $x_1$ have order 2. Since $x_1 \notin K$, $x_1 K = H$, ie $x_1 a$ has order 2, so $x_1 a x_1 a = e \implies x_1 a x_1 = a^{-1}$. Then $x_1 a^{-1}b^{-1} x_1 = (a^{-1}b^{-1})^{-1} = ba$ (since $a^{-1}, b^{-1} \in K$), but also $x_1 a^{-1}b^{-1} x_1 = x_1 a^{-1} x_1 x_1 b^{-1} x_1 = ab$, so $K$ must be abelian.
PS: It wasn't really necessary to use the efficient subgroup criterion, it would have worked just as well to say that for any $i$, $x_i x_j = x_i x_k \implies x_j = x_k$ and $x_j x_i = x_k x_i \implies x_j = x_k$ and directly deduce closure and inverses. It's just that that becomes a little messy.