96. (9th grade) Find the sum of all irreducible fractions with a denominator of 3, lying between the positive integers \(m\) and \(n\) (\(m<n\)).
96. We find the sum of fractions. There are a total of intervals between integers $n-m$, with two fractions in each interval.

$\left(m+\frac{1}{3}\right)+\left(m+\frac{2}{3}\right)+\left(m+\frac{4}{3}\right)+\left(m+\frac{5}{3}\right)+\ldots+\left(n-\frac{2}{3}\right)+\left(n-\frac{1}{3}\right)=$

$=m+m+\ldots+m+n+n+\ldots+n=m(n-m)+n(n-m)=n^{2}-m^{2}$.