Consider three of the $n$ points. The parity of the number of blue segments of the triangle with these points as vertices doesn't change while switching the keys. Since it is possible to make all three segments red, the number of blue segments in each triangle is even.

Let $P$ be one of the $n$ points. Let $A$ be the set of points connected to $P$ by red segments and $B$ be the set of points connected to $P$ by blue segments. Let $A_1, A_2 \in A$. So $PA_1$ and $PA_2$ are both red and thus $A_1A_2$ is red. Now consider $B_1, B_2 \in B$. Then $PB_1$ and $PB_2$ are both blue and thus $B_1B_2$ is red. Finally, consider $A \in A$ and $B \in B$. $PA$ is red and $PB$ is blue, so $AB$ is blue. Put $P$ in $A$. All this reasoning shows that segments in the same set are red and segments connecting points in different sets are blue.

Switching all points in set $A$ will make all segments red. Indeed, all segments in $A$ will change twice, one time from each of its edges, all segments connecting points from $\mathcal{A}$ and $\mathcal{B}$ will change once, turning from blue to red and segments in $\mathcal{B}$ won't change. This proves the first part.

For the second part, notice first that one needs to switch each point at most once. Let $|\mathcal{A}| = k$ and $|\mathcal{B}| = n-k$. If we switch $a$ points from $\mathcal{A}$ and $b$ points from $\mathcal{B}$ we change at most $a(n-k) + bk$ blue segments. Suppose without loss of generality that $k \le n-k \iff k \le \lfloor \frac{n}{2} \rfloor$. Then $k(n-k) \le a(n-k) + bk \le a(n-k) + b(n-k) \implies k \le a+b$. So the number of key presses is at most $k$ and, in the worst case, $\lfloor \frac{n}{2} \rfloor$. This number is needed to make all segments red if $|\mathcal{A}| = \lfloor \frac{n}{2} \rfloor$.