2010 AMC 12A Problem 22

The function $f$ is piecewise linear with breakpoints at $x=\tfrac1k.$ On the interval $\left[\tfrac1m,\tfrac1{m-1}\right]$ its slope is $$\begin{aligned} &\sum_{k=m}^{119}k \\ &\quad {}-\sum_{k=1}^{m-1}k=7140-(m-1)m, \end{aligned}$$ where $7140=\tfrac{119\cdot120}{2}.$ This slope is zero when $(m-1)m=7140,$ i.e. $m=85,$ so the minimum occurs at the right endpoint $x=\tfrac1{84}.$ There, terms with $k\le84$ contribute $\tfrac{84-k}{84}$ and terms with $k\ge85$ contribute $\tfrac{k-84}{84},$ so $$\begin{aligned} f\left(\tfrac1{84}\right) &= \frac{3486}{84}+\frac{630}{84} \\ &= 41.5+7.5=49. \end{aligned}$$ Thus, A is the correct answer.