A problem: Georgina calls a 992-element subset $A$ of the set $S = \{1, 2, 3, . . . , 1984\}$ a halfthink set if
• the sum of the elements in $A$ is equal to half of the sum of the elements in $S$, and
• exactly one pair of elements in $A$ differs by 1.
She notices that for some values of $n$(less than 1984), there are no halfthink sets containing both $n$ and $n+1$. Find the all possible values of $n$.
I call the values $n$ is halfthink if there are a halfthink set containing both $n$ and $n+1$ and I have tried this problem for easy cases:
Case I: $S=\{1,2,3,4,5,6,7,8\}$. We need find four numbers that the sum of them is 18. Notice that all even numbers, 2,4 and,6 are halfthink. 4 is obviously halfthink since we have 2+4+5+7=18; 2 is halfthink since we have 2+3+5+8=18; symmetrically, we can get 7+6+4+1=18, so 6 is halfthink. We can claim that if $n$ is halfthink, then so is $8+1-n$ and vice versa. Then all odd numbers are not half think. For example, if we choose 1 and 2, then we need to find two numbers left that the sum of them is 15(=18-3), however the only possible pair is 7 and 8. Similarly we can know that 3 is not halfthink, so symmetrically we can get 5 and 7 is not halfthink.
Case II: $S=\{1,2,3,4,5,6,7,8,9,10,11,12\}$. We need find six numbers that the sum of them is 39. Similarly all even numbers are halfthink, since 2+3+5+7+10+12=39, 1+4+5+7+10+12=39, 2+4+6+7+9+11=39, and the symmetrical property. For odd numbers, 3, 5, 7 and 9 are still not halfthink, however 1 and 11 is halfthink for 1+2+6+8+10+12=39.
...
So I guess that for $S=\{1,2,...,4m\}$, all even numbers, 2,4,..., and $4m-2$, are halfthink; only $2m-3, 2m-1, 2m+3, 2m+5$ are not halfthink and the odd numbers left are halfthink.
Is that true? How to prove that?