Gọi $S$ là số giá trị $m$ nguyên thuộc khoảng $\left( -20;20...

Vì $f(0)=48>0$ nên hàm số $y=\left| f(x) \right|$ đồng biến trên khoảng $\left( 0;2 \right)$ khi và chỉ khi $f'(x)\ge 0,\forall x\in \left( 0;2 \right)$
$\Leftrightarrow 8{{x}^{3}}-12(m+4){{x}^{2}}+6{{m}^{2}}{{x}^{{}}}\ge 0, \forall x\in \left( 0;2 \right)$
$\begin{aligned}
& \Leftrightarrow g(x)=4{{x}^{2}}-6(m+4){{x}^{{}}}+3{{m}^{2}}\ge 0,\forall x\in \left( 0;2 \right) \\
& \\
\end{aligned}$
$\Leftrightarrow \left[ \begin{aligned}
& \Delta '\le 0 \\
& \left\{ \begin{aligned}
& \Delta '>0 \\
& g(0)\ge 0 \\
& g(2)\ge 0 \\
& \left[ \begin{aligned}
& \dfrac{S}{2}\ge 2 \\
& \dfrac{S}{2}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& -3(-48-24m+{{m}^{2}})\le 0 \\
& \left\{ \begin{aligned}
& -3(-48-24m+{{m}^{2}})>0 \\
& {{m}^{2}}\ge 0 \\
& 3{{m}^{2}}-12m-32\ge 0 \\
& \left[ \begin{aligned}
& \dfrac{3m+4}{4}\ge 2 \\
& \dfrac{3m+4}{4}\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left[ \begin{aligned}
& m\le 12-8\sqrt{3} \\
& m\ge 12+8\sqrt{3} \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& \\
& 12-8\sqrt{3}<m<12+8\sqrt{3} \\
& \left[ \begin{aligned}
& m\ge \dfrac{6+2\sqrt{33}}{3} \\
& m\le \dfrac{6-2\sqrt{33}}{3} \\
\end{aligned} \right. \\
& \left[ \begin{aligned}
& m\ge \dfrac{4}{3} \\
& m\le -\dfrac{4}{3} \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& m\ge \dfrac{6+2\sqrt{33}}{3} \\
& m\le \dfrac{6-2\sqrt{33}}{3} \\
\end{aligned} \right.$
$\Rightarrow m\in \left\{ -19;-18;...;-2 \right\}$ $\bigcup \left\{ 6 ;7 ; 8 ;...; 19 \right\}$
Suy ra $S=32$. Vậy $S$ chia hết cho $4.$