Unique way to partition into two parts of equal weight A special case says it all ... Let $ w_1 < w_2 < \ldots < w_{12} $ be an increasing sequence of $12$ integers ("weights") such that the total weight $W=\sum_{k=1}^{12}w_k$ is even. 
Say that $I \subseteq \lbrace 1,2, \ldots ,12 \rbrace$ is an exact subset iff
the sum $\sum_{k \in I}w_k$ equals $\frac{W}{2}$. My question is : is there
a sequence for which $ \lbrace 1,2,5,7,10,12 \rbrace $ is exact and is
the only exact subset (up to complementation) ? 
 A: The answer is yes.  Consider the sequence 
$100, 200, 201, 202, 500, 601, 700, 701, 801, 1000, 1194, 1200.$
It is easy to see that $X=\{1,2,5,7,10,12\}$ and $Y=\{3,4,6,8,9,11\}$ are exact.  Moreover, we claim that they are the only exact subsets. To see this, note that for every subset $X'$ of $X$, the sum of the corresponding sequence values is divisible by $100$.  However, for every proper subset $Y'$ of $Y$, the sum of the corresponding sequence values is not divisible by $100$.  Thus, $X$ and $Y$ are the only exact subsets.  
A: Just an expansion on my comment. I will assume that exact sequences need to have half the number of indices of the whole sequence. Then a sequence is exact for some choice of weights if and only if it is exact for one weight. (This is already argued, both in my comment and in Tony Huynh's.)
The final question to be answered is when is a subsequence exact for some choice of weights. This is again very easy. A subset I of {1,...,2n} of size n obviously determines an increasing bijection $j_I$ between I and its complement. The subset I is an exact subsequence for some choice of weights if and only if the function $i \mapsto j_I(i)-i$ for $i \in I$ does not have constant sign.
