The k-th sect is defined as follows. Label the people i, i=1 to 20:

Sect(k)={0, 2, 7, 8, 11}+k (mod 21).
The set (with or without 21) is a golomb ruler:if s(i)-s(j)=n=s(l)-s(m), then i=l and j=m, that is, each difference can only be made one way.

This means that if two sects have a person in common, then s(i)+n=s(l) for some n. Then there can be only one overlap: if s(j)+n=s(m), then rearranging, we get the Golomb condition applying. 

There will always be an overlap because the Circle Golomb ruler ( or modular) is perfect: each number can be made in some way, as twice 5C2 is 20.

You can also express the solution as a Finite Projective Plane, like the Fano plane. Here’s a visual solution:
http://www.maa.org/editorial/mathgames/21lines.gif

The finite projective plane has the same definition as the problem.

N=3 SPOILER(?):

For the n=3 case, the solution is beautifully represented by the Fano plane. Each sect is a line, each point a person (or visa-versa).

I think the correct place to look for the n=5 solution will be the study of error-correcting codes, especially hamming codes. Eg. The Fano plane is hamming(7,4).

Try representing each sect as a binary number, with a 1 in position n if it contains person n, and a 0 otherwise (or the other way round). This will coincide with the hamming code in the n=3 case.