Let us call the number of ways of choosing a subset of N people such that there exist 3 people in the subset who are adjacent in the original set as $a_n$. Call such subsets good.
Let's try to find a recurrence for $a_n$.
Now, let us take any such subset of $N-1$ people, and consider that set with $N$ in it and not in it. Clearly both are distinct good subsets. So, this amounts to $2 a_{n-1}$ ways.
So, the elements not yet considered have $N$ in the triplet, and in fact the triplet $N-2, N-1,N$ should be the only one inside the subset, otherwise it has been calculated before. So, this amounts to $2^{n-4} - a_{n-4}$ ways, as we need the number of subsets of $N-4$ people not having any such triplet (Note that the element $N-3$ cannot be in the set).
So, we get the recurrence as $a_{n} = 2a_{n-1} + 2^{n-4} - a_{n-4}$ with initial conditions as $a_1=a_2=0, a_3=1, a_4=3$.
As regards to a general formula, the use of OEIS leads to this which tells us that the generating function is $G(x) = \frac{x^3}{(1-2x)(1-x-x^2-x^3)}$ and also gives a nicer looking recurrence: $a_n = 3a_{n-1} - a_{n-2} - a_{n-3} - 2a_{n-4}$