Why this question is the fourth question? I think it must be 1 or 2 .

Mathematicalx wrote: Why this question is the fourth question? I think it must be 1 or 2 . Can you show us your solution?

Summary of solution Let $A$ to place the balls such that $1,1,0,1,0,1,0,1....0,1,0$ cyclicly . Then its very easy to prove that $A$ can guarentee to take $2011$ balls in each case: $1)$ $B$ chooses the box which contains one of the two consecutive 1 balls. $2)$ $B$ chooses the box which contains 1 ball which is not consecutive with another balls. $3)$ $B$ chooses the box which contains no ball near the two consecutive 1 balls. $4)$ $B$ chooses the box which contains no ball and isnt near the two consecutive 1 balls. Finally its very easy to prove that $A$ cant guarentee to take $2012$ balls.

EDIT: I misunderstood the problem badly. I thought, for example, with the boxes $a_1, a_2, \ldots, a_{2n+1}$ and B taking box $a_{2n+1}$, A can only take either the odd-numbered boxes or the even-numbered boxes. This is also a good variant of the problem it seems, since I've been thinking like that for a few days...

I agree with you, problem is pretty. But i think its not convenient (too easy) for problem 4 .

Mathematicalx wrote: Summary of solution Let $A$ to place the balls such that $1,1,0,1,0,1,0,1....0,1,0$ cyclicly . Then its very easy to prove that $A$ can guarentee to take $2011$ balls in each case: $2)$ $B$ chooses the box which contains 1 ball which is not consecutive with another balls. If he takes 1 from 6th box $A$ will not have an ability to take 2011 balls

If he takes 1 from 6th box A will not have an ability to take 2011 balls.İ agree with u.bigant146

This is the official solution Let's first show that Player A can collect 1342 balls. To do this, Player A places 9 boxes around the circle, numbers them 1 to 9 clockwise, and puts 671 balls into each of the boxes numbered 1, 4, and 7. It is clear that in every selection, Player B can choose two non-empty boxes for Player A. Now, let's prove that it is impossible for Player A to guarantee collecting more than 1342 balls. Suppose that for some number 2m + 1, the balls can be distributed in such a way that this is possible. Number the boxes 1 to 2m + 1 clockwise. Only two of the boxes not chosen by Player A will be adjacent to each other. Without loss of generality, let the numbers of these two boxes be 2m + 1 and 1. Let the boxes chosen by Player A be called type A boxes, and the unchosen ones be called type B boxes. According to our assumption, the total number of balls in the type A boxes will exceed 1342. Therefore, there will be some number t such that the total number of balls in the type A boxes numbered 1 < i ≤ t will be at least 672, and the total number of balls in the type A boxes numbered t ≤ i < 2m + 1 will also be at least 672. In this case, if Player B chooses the t-numbered box at the start, Player A cannot select all the type A boxes numbered 1 < i ≤ t or t ≤ i < 2m + 1, and as a result, the total number of balls in the boxes chosen by Player A will be at most 1341. This contradiction completes the solution to the problem.