Will it even end?

Relevant wiki: Polynomial Inequalities Problem Solving - Basic

$\left(\dfrac{4}{x-3}-\dfrac{4}{x-2}\right)+\left(\dfrac{1}{x-1}-\dfrac{1}{x-4} \right)< \dfrac{1}{30}$ $\implies \dfrac{4}{\underbrace{(x-2)(x-3)}_{(\color{#D61F06}{x^2-5x+4})+2}}-\dfrac{3}{\underbrace{(x-1)(x-4)}_{\color{#D61F06}{x^2-5x+4}}}<\dfrac1{30}$

Put $\color{#D61F06}{x^2-5x+4}=t$

$\implies \dfrac4{t+2}-\dfrac3{t}<\dfrac1{30}$ $\implies \dfrac{t-6}{t(t+2)}<\dfrac1{30}$ $\implies \dfrac{t^2-28t+180}{t(t+2)}>0$ $\implies \dfrac{(t-10)(t-18)}{t(t+2)}>0$ Put $t$ back and factorise further to obtain:

$\implies \dfrac{(x+1)(x+2)(x-6)(x-7)}{(x-1)(x-4)\left(x-2\right)\left(x-3\right)}>0$

Plot a number line to see solution set as:-

$x\in(-\infty,-2)\cup(-1,1)\cup\left(2,3\right)\cup(4,6)\cup(7,\infty)$

Hence,

$-2-1+1+2+3+4+6+7=\boxed{20}$

Call the given function $f(x)$ .

Notice that $f(x)=f(5-x)$ , which gives $h = 5 - a$ , $g = 5 - b$ , $f = 5 - c$ and $e = 5 - d$ .

So $a + b + c + d + e + f + g + h = (a + h) + ( b + g) + (c + f) + (d + e) = 5 + 5 + 5 + 5 = \boxed{20}$