I thought I had it but I guess not.

The two disjoint collections must partition the set of 32 numbers. That is, the two disjoint collections must each add up to 60.

Let Si = <Sum k=1 to k=i> a_k

by the pigeonhole principle,two of the Si's must be congruent mod30.
For some i and j with 1<= i<j <=32

we have a_k+1 + a_k+2 + ... + a_j is a multiple of 30.

Therefore it'sums to 30,60 or 90. If it sums to 90, then taking the complement we get a subset which sums to 30. We conclude that a sum of 30 or 60 is attainable.

Given a set of n+2 integers that sums to 4n, and none of the integers exceeds 2n, then we can always find two disjoint sets each of which sums to 2n.

Sol'n : I will proceed using mathematical induction.

The statement is obviously true for n=1 and can be shown to be true for n=2.

Moreover, for n=k=2 the result below can be shown to be true.

That is, for n=k, given any set of k+2 integers that sums to 4k+3, we can always find a subset that sums to 2k+1; given any set of k+2 integers that sums to 4k+2, we can always find a subset that sums to 2k; given any set of k+2 integers that sums to 4k+1, we can always find a subset that sums to 2k-1.

We need to prove that given a set of (K+1) + 2 integers that sums to 4(k+1), and none of the integers exceeds 2(k+1), then we can always find two disjoint sets each of which sums to 2(k+1)

proof: If one of the integers is 2(k+1), then we're done. Additionally, if one of the integers is 2(k+1) - 1, then we're done since
2(k+1) - 1 + 2(k+2) = 4k+5 > 4k + 4

Therefore we can assume that all of the integers is less than or equal to 2*k.

Since 4(k+1)/(k+1+2) < 4 for all values of k, then one of the (k+1) + 2 integers must be 1, 2 or 3.

If it is 1, then of the remaining K+2 integers that are less than or equal to 2k, there is a subset that sums to 2k+1

If it is a 2, then of the remaining k+2 integers that are less than or equal to 2k, there is a subset that sums to 2k.

If it is a 3, then of the remaining k+2 integers that are less than or equal to 2k, there is a subset that sums to 2k-1.

We can check that 1 + 2k+1 = 2k+2, 2 + 2k= 2k+2 and 3+2k-1 = 2k+2

completing the induction.

Here is the problem.

If the above is true for general k, then the problem is solved.

I can't prove this in general.

I'm optimistic that a proof by mathematical induction might work.

The 60/60 problem is quite interesting, but I don't know whether there's a short proof. I've more or less proved it by contradiction: obviously 60 can't be used, 59 can't be used (because at least one 1 is then required), 58 can't be used (because at least a 2 or two 1s are then required), 57 can't be used, etc., down to 4. Then 3, 2, 1 are obvious. Messy.

I'd say that a better argument may be to look for the largest integer.

2*14 + 3 + 29 = 60.

Thus the induction can't possibly work.

Order the numbers from least to greatest.

a1<=a2<=a3<=...<=a32

Place a32 in set A and a31 in set B. If they are equal, then place a30 in set A. Otherwise if a32 and a31 are unequal, then we place a30 in set B and we continue adding elements to set B until the sum of the elements is greater than or equal to set A. Once this occurs, we continue adding elements to set A and the process is continued.

First we need to show that if the sum of the elements of A is S, then the sum of the elements from set B cannot jump from something less than S to something greater than 60 for 30 <= S <= 57

since (61 + S)/(61 - S) + 120 - (61 + S) < 32

This shows that the sum of the elements in either set must eventually reach 57. We can eliminate 58-62 since this would imply that the sum of the 32 numbers is at least
4*32 > 120, a contradiction. Therefore we can conclude that the game will terminate with the score 59-61 or 60-60.

WLOG, suppose set B is 61 and set A is 59.

RTP : We can always find a subset from set B that adds to k+1 and a subset from set A that adds to k for 2 <= k <= 57

Proof: Suppose the above claim is untrue.

case 1: The score is 58-58 and the last two elements are 3 and 1 making the final score 59-61. Retracing our steps backwards we sum the elements from least to greatest such that the sum of the 32 numbers is a minimum. Of course, without the above restrictions the sum of these numbers must add to 120. To minimize the sum of these elements we require one selection from set A, then another selection from set B, then another from set A and so on until we have 16 members in each set.

The sum of the 32 numbers must be at least 1 + 3*19 + 4 + 6*11 > 120, a contradiction.

case 2: The score is 59-58 and the last element is a 3 making the final score 59-61. As before, we find a lower bound for the sum of these numbers which occurs when we've selected 16 members in each set. The sum of the 32 numbers must exceed 120 which we can verify since

3*19 + 4 + 5*12 > 120

case 3: The score is 59-59 and the last element is a 2. The sum of the 32 numbers must be at least

2*29 + 3 + 30*2 > 120, a contradiction.

In all three cases we've reach a contradiction which shows that a trade off must occur to reach 60-60. Simply take the sum of the elements from set B that adds to k+1 and trade it to set A for the sum of the elements from A that adds to k.

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