Proof. Let m be the number of people in some group. Then the internal monitoring leaves the group with a maximum of m(n-(m-1)) members of other groups it can still monitor, i.e. mi(n-(mi-1)) is for every i an upper bound for the sum of all mj, j≠i.
Considering the smallest m, we find that it can not be smaller than the sum of the other m divided by n and in particular that there can not be more than n+1 groups.
And considering the biggest m, we find that it can not exceed n+1.
Actually, Σjmj=mi(n+2-mi) for every i, is being solved by
mi=n/2+1, where i=1, ..., n/2+1,which is a good candidate for the maximal sum. In general the solution to mi(n+2-mi)=mj(n+2-mj) for some pair i≠j, 0<mi<n+2, is
n/2+1±((n/2+1)²-mi(n+2-mi))0.5=n/2+1±(mi²-(n+2)mi+(n/2+1)²)0.5=n/2+1±(n/2+1-mi),that is all mi must take one of two equidistant values to n/2+1. Let m and m' be those two values, i.e. m'=n+2-m and m=n+2-m'. Let a be the number of groups of size m and b be the number of groups of size m'. Then am+bm'=mm' and this value has to be maximal. But m(n+2-m) is maximal for m=n/2+1.
Hence the actual limit to the size of the set of people who monitor each other should be (n/2+1)², but my point here has little to do with the exact value. It is this:
The monitoring of a potentially infinite set of people must be structured and effective structures are at least implicitly hierarchical, in so far that they have a center.