Haha jokes on you, I managed to solve the problem. Here's a (slightly) more algebraic approach:

We can simplify the compound inequality as follows:

$\begin{aligned} 2x&<x-4&\le 3x+8\\ x+4&<0&\le 2x+12 \end{aligned}$

For this to be true, we must have $x+4 < 0$ and $2x + 12 \ge 0$ . The answer must be the union of $x < -4$ and $x \ge -6$ , or $-6 \le x < -4$ . The only integer solutions to this inequality are $-5$ and $-6$ , and the sum of the two is $\boxed{-11}$ .

Thanks again for the featured problem, and the Mathathon 2020! It was a lot of fun!

As seen from the graph, the only integer values that satisfy this equation are $-5 \ and \ -6$

$-5 + -6 = \textsf{\boxed{-11}}$

$2x<x-4\leq3x+8$ $\Rightarrow 2x<x-4\quad x-4\leq3x+8$ $\therefore x<-4\quad x\geq-6$ As $x$ is also an integer $\therefore x\in \{-5,-6,-7...\}\cap\{-6,-5,-4...\}=\{-5,-6\}$ $-5-6=\boxed{-11}$