jmc

Let $$f(x)=|x-1|+|2x-1|+|3x-1|+\cdots +|119x-1|.$$If $x\le \frac{1}{119},$ then $$f(x)=-(x-1)-(2x-1)\cdots -(119x-1).$$If $\frac{1}{m}\le x\le \frac{1}{m-1},$ for some positive integer $2\le m\le 119,$ then $$f(x)=-(x-1)-(2x-1)\cdots -((m-1)x-1)+(mx-1)+\cdots +(119x-1).$$If $x\ge 1,$ then $$f(x)=(x-1)+(2x-1)+\cdots +(119x-1).$$Thus, the graph is linear on the interval $x\le \frac{1}{119}$ with slope $-1-2-\cdots -119,$ linear on the interval $\frac{1}{m}\le x\le \frac{1}{m-1}$ with slope $$-1-2-\cdots -(m-1)+m+\cdots +119,$$and linear on the interval $x\ge 1$ with slope $$1+2+\cdots +119.$$Note that $$\begin{array}{rl}-1-2-\cdots -(m-1)+m+\cdots +119& =-\frac{(m-1)m}{2}+\frac{(m+119)(120-m)}{2}\\ & =-m^2+m+7140\\ & =-(m+84)(m-85).\end{array}$$Thus, $f(x)$ is minimized on the interval $\frac{1}{85}\le x\le \frac{1}{84},$ where it is constant, and this constant is $$(85-1)-(119-85+1)=\boxed{49}.$$