smc

If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$, where $m$ is any integer from $1$ to $119$, inclusive: $|mx-1|=0⟹mx=1⟹x=\frac{1}{m}$. The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$, since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$. The sum of the slopes at $x=\frac{1}{m}$ is $$\begin{array}{rl}& \sum _{i=m+1}^{119}i-\sum _{i=1}^{m}i\\ & =\sum _{i=1}^{119}i-2\sum _{i=1}^{m}i\\ & =-m^2-m+7140\end{array}$$ Now we want to minimize $-m^2-m+7140$. The zeros occur at $-85$ and $84$, which means the slope is $0$ where $m=84,85$. We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49}$. You can also think of the slopes playing 'tug of war', where the slope of each absolute function upon passing its $x$-intercept is negated, positively tugging on the remaining negative slopes. The sum of the slopes is $1+2+3+4\dots 119={\sum }_{m=1}^{119}m=\frac{119\cdot 120}{2}=60\cdot 119=7140$ So we need to find the least integer $a$ such that $1+2+3+\dots a={\sum }_{n=1}^{a}n=\frac{a(a+1)}{2}\ge \frac{7140}{2}=3570:$ $$a(a+1)\ge 7140⟹a^2+a-7140\ge 0\to a=84\text{ exactly!}$$ This "exactly" means that the slope is ZERO between the whole interval $x\in \left(\frac{1}{85},\frac{1}{84}\right)$. We can explicitly evaluate both to check that they are both equal to the desired minimum value of $f(x)$: $$\frac{84+83+\cdots +2+1+1+2+\cdots +33+34}{85}=\frac{84(85)/2+34(35)/2}{85}=\frac{85(14+84)/2}{85}=49$$ $$\frac{83+82+\cdots +2+1+1+2+\cdots +34+35}{84}=\frac{83(84)/2+35(36)/2}{84}=\frac{84(15+83)/2}{84}=49$$ Thus the minimum value of $f(x)$ is $49$.