Finding values of $t$, for 0, 1, 2, or > 2 roots of: $\sum_{k=1}^{50} A_k~ |x-k|= t.$ Find values of $t$, if the equation: $$\sum_{k=1}^{50} A_k~ |x-k|= t, A_1=3, A_2=4, A_3=5, A_{k \ge 4}=1,$$ has 0, 1, 2 or more than 2 roots.
This question is inspired by a recent interesting question:
How many solutions has this equation with a parameter?
But the present question contains additional three new features and thoughts.
I hope that this question will also be found interesting for solutions. Also notice four tags I have chosen for this question.
 A: Let us denote the given equation as  $$f(x) = \sum_{k=1}^{50} A_k ~|x-k|=t, A_1=3, A_2=4, A_3=5, A_{k\ge 4}=1.....(1).$$ Then, $f(x)$ is nothing but 59 times the mean deviation of a sequence of 59 natural numbers about $x$. The sequence is:  1, 1, 1, 2, 2, 2, 2, 3 , 3 , 3 , 3 , 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,......, 48, 49, 50.  It is known that mean deviation is the least when measured about the median. The median of this data of 59 numbers is [59/2]+1=30$^{th}$ element of the said sequence which is 21. Therefore $f(x)$ will admit the local minimum at $x=21$. As $f(21)=814$, if $t<814$, the Eq. (1) has no roots. If $t=814$, it has exactly one root. if $t>814$, it has two roots. This equation cannot admit more than two roots.
Graphically, $y=f(x)$ is piece-wise continuous, open polygon diverging to infinity on both  sides of  the  single local minimum at $x=21$ and $f_{min}=f(21)=814.$. Then the line $y=t$ can touch/cut the polygon at most at 2 points.

