Find all positive integers n, for which the numbers in the set S = {1, 2, . . . , n} can be colored red and blue, with the following condition being satisfied: the set S × S × S contains exactly 2007 ordered triples (x, y, z) such that (i) x, y, z are of the same color and (ii) x + y + z is divisible by n.
The answer is n = 64  and n = 84.

Let the numbers 1,2, ...n be colored red and blue. They are partitioned into sets of Red and Blue and denoted by R and B respectively. The size of R = r and the size of B = n - r. The triple (x, y, z) belonging to triple S x S x S is monochromatic if x, y, and z have the same color and bichromatic otherwise.(x,y,z) is divisible if  x+y+z is divisible by n.
Here the claim is there are exactly r^2-rb+b^2 divisible monochromatic triples
For any pair (x,y) that belongs to  S x S , there exists  a unique z for the chosen x,y that belongs to S such that the triple is divisible. There are exactly n ^2 divisible triples. Moreover if the triple is bichromatic, then x, y, z are either blue or two red or vice versa. In both those cases, there is exactly one pair that belongs to the set R x B.. We assign such pair to triple (x, y, z).
Consider any pair (x,y) belonging to RxB, and denote z = z for  a given x,y. Since x != y, the triples x,y,z in any order is distinct and x,y is assigned to each of them. Each pair in RxB is assigned exactly three times.
Therefore, the number of bichromatic divisible triples is three times the number of elements in RxB and the number of monochromatic ones is n^2 - 3rb where n = r + b and can be written as r^2-rb+b^2
To find all values of n where the desired coloring is possible, we have to find all n, for which there exists a composition n = r + b and r^2 -rb + b^2 = 2007.
Solving for these two equations we get (r,b)=(51,18) or (r,b)=(51,33) and the possible values of n = 51+18=69 and n = 51+33=84.