Numbers on a circle: how many arc sums can be positive? There are $n$ real numbers, $a_1,\dots,a_n$, arranged on a circle. Given a fixed integer $k<n$, let $S_i$ be the sum of the $k$ adjacent numbers starting at $a_i$ and counting clockwise, like this (illustrated for $k=3$): 

The sum of all $n$ numbers is $0$. What is the largest possible number of strictly-positive $S_i$?
Here are some lower bounds:


*

*If $k\leq n/2$, then at least $n-k$ sums can be positive, e.g. when $a_1 = \cdots = a_{n-1} = 1$ and $a_n = -(n-1)$. There are $n-k$ sums that do not contain $a_n$, and they are positive.

*If $k\geq n/2$, then at least $k$ sums can be positive, e.g. when $a_1 = \cdots = a_{n-1} = -1$ and $a_n = +(n-1)$. There are $k$ sums that contain $a_n$, and they are positive.


And here are some upper bounds:


*

*If $n$ is even and $k=n/2$, at most $n/2$ sums can be positive. This is because, if a certain sum $S_i$ is positive, then its complement sum $S_{i+k}$ must be negative (since $S_i+S_{i+k}=0$).

*If $k=n/3$, then at most $2n/3$ sums can be positive, since if $S_i$ and $S_{i+k}$ are positive, then $S_{i+2k}=-(S_i+S_{i+k})$ must be negative.

*Similarly, if $k=n/a$ for some integer $a$, then at most $(a-1)n/a$ sums can be positive.


What is the general answer as a function of $k$?
 A: This isn't a complete answer (no upper bounds), but some more general constructions and remarks that make it too long for a comment.
As joriki pointed out, we can assume $n \ge 2k$.
The first construction is due to Aravind (see the comment above).  Suppose $n = qk + r$, $q \ge 2$ and $1 \le r \le k-1$ (if $r = 0$, then $k | n$, and then we know there can be $n - k$ positive sums, and this is best possible).  Now for $1 \le i \le n$, set
$ a_i = \begin{cases} k - 1 + \frac{r}{q} & \mbox{if } k | i, \\ -1 &\mbox{otherwise.}\end{cases}$  We then have $\sum_{i=1}^n a_i= 0$.
Any sum containing one of the $a_{jk}$ is positive, so the only negative sums are $S_{qk+1}, S_{qk+2}, ..., S_{qk+r}$.  Thus there are $n-r$ positive sums.  If $r = 1$, this is clearly optimal.
Note also that any placement of the $q$ positive terms, provided they are at least $k$ apart, works, since all $qk = n-r$ sums involving the positive terms will be positive.
When $r > \frac{k}{2}$, there is a construction that does better: it gives $n - k + r$ positive sums instead.  If $r = k-1$, this is again best possible.  For $1 \le i \le n-1$, set $a_i = \begin{cases} - \left(k-1- \frac{1}{2q} \right) &\mbox{if } k|i, \\ 1 &\mbox{otherwise.}\end{cases}$  Let $a_n = -\left(r - \frac12 \right)$.  We again have $\sum_{i=1}^n a_i = 0$.
Any sum involving only one negative term is positive, hence the only negative sums have two negative terms.  The only such sums have both $a_{qk}$ and $a_n$, and are $S_{n-k+1}, S_{n-k+2}, ..., S_{n-r}$ (recall that $qk = n-r$).  Hence there are $k-r$ negative sums, and thus $n-k+r$ positive sums.
Finally, I'd like to point out that if the answer had indeed been $n-k$, this would have proven the Manickam-Miklos-Singhi conjecture.  The MMS conjecture has been the focus of much recent research.  It claims that if $n \ge 4k$, then for any collection of $n$ numbers summing to $0$, there are at least $\binom{n-1}{k-1}$ subsets of size $k$ with non-positive sum.  This can be achieved, for example, by taking one number to be $-(n-1)$, and the others to be $1$.  The difficulty is in proving that there cannot be fewer.
This is known to be true if $k | n$, or if $n \ge 10^{46} k$ (a result of Pokrovskiy (paper)).  The condition $n \ge 4k$, or something similar, is necessary, since counterexamples are known for smaller $n$ (e.g. $n = 3k+1$).
Now suppose in this cyclic version, we could never have more than $n-k$ positive sums, so always had at least $k$ non-positive sums.  If we take a random cyclic ordering of the $n$ elements, every cyclic $k$-set is a uniformly random $k$-set.  Hence if at least $k$ have non-positive sum, that implies that at least a $\frac{k}{n}$-proportion of all $k$-sets, or $\binom{n-1}{k-1}$ $k$-sets, have non-positive sum, proving the conjecture.
Sadly, however, the above examples show that this cannot be used to prove the Manickam-Miklos-Singhi conjecture (at least not without additional work).
