Solution:

The function $f$ is piecewise linear with breakpoints at $x=\tfrac1k.$ On the interval $\left[\tfrac1m,\tfrac1{m-1}\right]$ its slope is $$\sum_{k=m}^{119}k-\sum_{k=1}^{m-1}k=7140-(m-1)m,$$ 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 $$f\left(\tfrac1{84}\right)=\frac{3486}{84}+\frac{630}{84}=41.5+7.5=49.$$

Thus, A is the correct answer.