The answer is 4. Let rectangle is ABCD, and E is midpoint of AB, F is midpoint of CD. 4 ball is sufficient) P:intersection of AC and DE Q:intersection of AC and BD R:intersection of AF and BD S:intersection of AF and DE We place 4 ball on P,Q,R,S, then satisfy condition. 3 ball is not sufficient) We suppose that we can place 3 ball α,β,Υ which satisfy condition.Then 2 pockets are on the line αβ, and the other 2 pockets are on the line βΥ, and the other 2 pockets are on the line Υα.A and F or A and C are on the same side line. Case1)A and F are on the same side line If E and B are on the same side line, 2 vertices are on AF and 2 vertices are on BE, which is absurd.Similarly, E and C are on the same side line, which is absurd.If E and D are on the same side line,B and C are on the same side line.Similarly, this is absurd. Case2)A and C are on the same side line Similarly Case1, E and F are on the same side line, B and D are on the same side line.1 vertice must be the intersection(=P) of AC and BD.1 vertice must be on EF which is not P.1 vertice must be on BD which is not P.1 vertice must be on AC which is not P.This is absurd. From Case1 and Case2, 3 ball is not sufficient. Therefore the answer is 4.