Let $\displaystyle w(j,h)=2\sum_{k=1}^{j-1}(-1)^{k+1}\left\lceil\frac{kh}{j}\right\rceil+h-1$, prove the following conjecture: $\displaystyle \sum_{j=1\ odd}^{h-1}w(j,h)>0$.
For the alternating sum $\displaystyle S(l,n)=\sum_{i=1}^{l-1}(-1)^{i+1}\left\lceil \frac{in}{l}\right\rceil $, I'd like to get some inequality for a lower bound that works for the above conjecture.
Assuming $h$ is an odd prime, then $(j,h)=1$, the remainders of $kj$ mod $h$ are always different. Let $j=2n+1$, by setting the negative terms to have the greatest remainders and the positive terms to have the smallest remainders, we get the following inequality:
$w(j,h)\geq 2(\sum_{k=1}^{n} \frac{(2k-1)h}{j}-\frac{\sum_{i=2n+1}^{2n}i}{j}+\frac{\sum_{i=1}^{n}i}{j}-\sum _{k=1}^n\frac{2kh}{j}=\frac{h}{2} -\frac{j}{2}-\frac{1}{2j}.$
But this bound is not sufficient for the above conjecture. Is there any way that we can arrange the remainders to get a better bound?