Another one of those interesting number theory questions:
Now if $\displaystyle n$ is even, say $\displaystyle 6$ then $\displaystyle 1,2,3,4,5,6$ any two adjacent ones you take we have one is even and one is odd. Thus $\displaystyle \gcd (a,n) , \gcd (a+1,n)$ cannot be both 1 because at least one of them is at least 2.
For the converse consider $\displaystyle 1,2,...,n$ and pick the first two and so $\displaystyle \gcd(1,n) , \gcd(2,n)$ we know that $\displaystyle \gcd(1,n)=1$ but since $\displaystyle F(n)=0$ it means $\displaystyle \gcd (2,n) \not = 1$ (otherwise $\displaystyle F(n)\not = 0$) and so it has to be $\displaystyle \gcd(2,n) = 2$ which means $\displaystyle 2|n$.