Partition the set into $n-1$ subsets of consecutive pairs of integers. See that the first can only pair with the rest of $n-3$ pars of subsets and the second one with only $n-4$ of these going all the way to 1. So, the total no. of perfect subsets is $\frac{(x-2)(x-3)}{2}$.

The answer should be $(n-3)(n-2)/2$. And this problem seems to be from the JBMO 2018 Shortlist, C1.

S=${/a,a+1,b,b+1}/$ so there are $n-2C2$=$(n-2)(n-3)/2$answers

So we can see that if $x-1 \in S$ then we have that $(x-1)+1\in S$ and the same goes for $x+1 \in S$, then we have that $(x+1)-1\in S$ So now we see that our subsets are of the following forms: $1)\{x,x+1,y,y+1\}$ $2)\{x,x-1,y,y+1\}$ $3)\{x,x+1,y,y-1\}$ $4)\{x,x-1,y,y-1\}$ But notice when we set $y \rightarrow k+1$ then the subset $3$ becomes the subset from $1$ Also notice that when we set $y \rightarrow k+1$ and $x \rightarrow j+1$, then the subset $4$ becomes the subset from $1$ Also notice that when we set $x\rightarrow j+1$ the the subset from $2$ becomes the subset $1$ So now we must calculate the number of subsets of the form of $\{x,x+1,y,y+1\}$ So now if we take into account that we have symmetry in the set, we get that the number of perfect subsets is $\frac{(n-2)(n-3)}{2}$