NMO Selection Tests for the Junior Balkan Mathematical Olympiad

If 1 is a separator, then we can not have any other separators than 1 and $n-1$.
For $n=1$ we can not have any separators.
For $n=2$, only 1 can be a separator (if the two numbers are equal).
For $n=3$ only 1 and 2 can be separators (for example, in the case of the numbers 1, 2, 3; $1+2+3=3$).
For $n=4$ we can have at most two separators, 1 and 3, for example in the case of the numbers 1, 2, 3, 6, with $1+2+3=6$.
We prove that, for $n \ge 5$, the maximum number of separators is $n-3$, namely in the case when all the numbers 2, 3, ..., $n-2$ are separators.
This maximum is achieved, for example, for the numbers
* $1, 1, 1, 1, 2, 2, 4, 4, \dots, 2^{k-2}, 2^{k-2}$ if $n=2k$
Indeed, when $n=2k$, we have $2S = 2^k$, hence $S = 2^{k-1}$ which can be written
$$S = 2^{k-2} + 2^{k-2} = 2^{k-2} + 2^{k-3} + 2^{k-3} = \dots = 2^{k-2} + 2^{k-3} + \dots + 2 + 1 + 1.$$

Also, when $n=2k+1$, we have $2S = 2^{k+1}$, hence $S = 2^k$ and we can write
$$S = 2^{k-1} + 2^{k-1} = 2^{k-1} + 2^{k-2} + 2^{k-2} = \dots = 2^{k-1} + 2^{k-2} + \dots + 2 + 1 + 1.$$