I interpret the question as, you can arrange the nobs in 2 dimensions however you like, and draw a line through them where one side is "in front" and the other is "behind". Call this line the front/back line.
At most 3 Nobs can be arranged in every possible set of preferences for those Nobs.
4 Nobs does not work: Suppose you have Nobs A, B, C and D, where A likes B, B likes C, C likes A, and D likes nobody. The only possible way to arrange A, B, and C is in a triangle where the front/back line of each Nob passes between the other 2. These three lines form an inner triangle, and inside of this triangle is the only place nobody is looking, and therefore the only place D can be. But, there's no way to draw a line passing through this triangle that does not pass between any 2 of A, B, and C, ie. a line that would allow D to see nobody, because this triangle is inside of the triangle formed by A, B, and C. So this set of preferences, with 4 Nobs, can't be satisfied.
3 Nobs does work: If you arrange 2 Nobs so that their front/back lines intersect, and then face them such that they are happy with the other one, you can place the third Nob in whichever of the 4 regions formed by these lines would satisfy the other 2 Nobs, and then draw a front/back line either between the first 2 Nobs (if Nob C likes only one) or not between the first 2 Nobs (otherwise), and you will be able to face Nob C along that line such that they are satisfied.
I strongly suspect that if you generalize the problem to N dimensions (eg. you can hang Nobs in the air looking slantwise),
N+1 Nobs is the maximum you can safely accommodate.