Et tu, Brute ....

Configure $20$ points in a circular pattern separated by equal distances. These points represent the $20$ board members. With these points we can form $\binom{20}{3} = 1140$ distinct triangles.

Now represent enemy pairings as a red line joining two points and friendships as a blue line joining two points. Now let the number of possible $3$ -person subcommittees that are not either all friends or all enemies be $N$ . Each of these $N$ triangles has $2$ of its vertices at the end of both a red and a blue line for a total of $2N$ vertices.

Now each of the $20$ points lies at one end of $6$ red lines and $13$ blue lines. So each of the points will be part of $6*13 = 78$ triangles which have sides which are not all the same color. This is the case for each of the $20$ points, and these $20*78 = 1560$ triangle vertices correspond to the $2N$ vertices discussed above. Thus

$2N = 1560 \Longrightarrow N = 780$ .

But the desired number of "all friend or all enemy" subcommittees will just be the complement $N$ , so the solution is $1140 - 780 = \boxed{360}$ .