Let $(G,\cdot)$ be a group of order $2n$ with $n$ elements of order $2$. Prove $n$ is odd and $G$ has an abelian subgroup of order $n$. 
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.
 A: 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.
A: Suppose that $|G| = 2n$ and that the number of elements of order $2$ in $G$ is $n$.
You can see that the number of $x \in G$ such that $x^2 = 1$ is even (for a proof, partition elements of $G$ into sets $\{x,x^{-1}\}$). This implies that the number of elements of order $2$ in $G$ is odd, so $n$ is odd.
Now consider the regular permutation representation of $G$, that is, $G$ acting on itself by left multiplication). In this action, an element $x \in G$ of order $2$ acts with cycle structure $$(x_1 x_2) (x_3 x_4) \cdots (x_{2n-1} x_{2n}).$$
In particular $x$ acts as an odd permutation, so $G$ has a normal subgroup $H$ of index $2$. (In general if $G \leq S_n$ and $G \not\leq A_n$, then $G \cap A_n$ is a normal subgroup of $G$ with index $2$.)
Let $x \in G$ have order $2$ and let $H \leq G$ be a subgroup of index $2$. Now $H$ has no elements of order $2$ since $|H| = n$ is odd, so every element of $xH$ has order $2$.
Then for all $h \in H$, we have $$xhx^{-1} = xhx = (xh)^2h^{-1} = h^{-1}.$$ So conjugation by $x$ gives an automorphism $h \mapsto h^{-1}$ of $H$. The inverse map is an automorphism if and only if the group is abelian, so $H$ must be abelian.
