What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$?
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$
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: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$.
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{align*}&\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{align*}
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 $\boxed{49\ \textbf{(A)}}$.
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 $1+2+3+\dots a=\sum_{n=1}^an=\frac{a(a+1)}{2}\ge \frac{7140}{2}=3570:$
\[a(a+1)\ge 7140\implies a^2+a-7140\ge 0\rightarrow a=84\text{ exactly!}\]
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 $f(x)$:
\[\frac{84+83+\dots+2+1+1+2+\dots+33+34}{85}=\frac{84(85)/2+34(35)/2}{85}=\frac{85(14+84)/2}{85}=49\]
\[\frac{83+82+\dots+2+1+1+2+\dots+34+35}{84}=\frac{83(84)/2+35(36)/2}{84}=\frac{84(15+83)/2}{84}=49\]
Thus the minimum value of $f(x)$ is $49$.