Let x_{1} \leq x_{2} \leq \cdots \leq x_{100} be real numbers such that \left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{100}\right|=1 and x_{1}+x_{2}+\cdots+x_{100}=0. Among all such 100-tuples of numbers, the greatest value that x_{76}-x_{16} can achieve is \frac{m}{n}, where m and n are relatively prime positive integers. Find m+n.