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.