Lời giải chi tiết:

Xét hàm số \(g\left( x \right) = \dfrac{1}{4}{x^4} - 14{x^2} + 48x + m - 30\) có TXĐ \(D = \mathbb{R}\).

Ta có \(g'\left( x \right) = {x^3} - 28x + 48 = 0 \Leftrightarrow \left[ \begin{array}{l}x =  - 6 \notin \left[ {0;2} \right]\\x = 4 \notin \left[ {0;2} \right]\\x = 2 \in \left[ {0;2} \right]\end{array} \right.\)

TH1: \(\mathop {\max }\limits_{\left[ {0;2} \right]} y = y\left( 0 \right) \Leftrightarrow \left\{ \begin{array}{l}y\left( 0 \right) \le 30\\y\left( 0 \right) \ge y\left( 2 \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m - 30} \right| \le 30\\\left| {m - 30} \right| \ge \left| {m + 14} \right|\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m - 30} \right| \le 30\\{\left( {m - 30} \right)^2} - {\left( {m + 14} \right)^2} \ge 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l} - 30 \le m - 30 \le 30\\ - 44\left( {2m - 16} \right) \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}0 \le m \le 60\\m \le 8\end{array} \right.\,\, \Leftrightarrow 0 \le m \le 8\).

TH2: \(\mathop {\max }\limits_{\left[ {0;2} \right]} y = y\left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}y\left( 2 \right) \le 30\\y\left( 2 \right) \ge y\left( 0 \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m + 14} \right| \le 30\\\left| {m + 14} \right| \ge \left| {m - 30} \right|\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m + 14} \right| \le 30\\{\left( {m + 14} \right)^2} - {\left( {m - 30} \right)^2} \ge 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l} - 30 \le m + 14 \le 30\\44\left( {2m - 16} \right) \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 34 \le m \le 16\\m \ge 8\end{array} \right.\,\, \Leftrightarrow 8 \le m \le 16\).

\( \Rightarrow S = \left\{ {0;1;2;...;16} \right\} \Rightarrow \) Tổng giá trị các phần tử của tập hợp \(S\) bằng: \(\dfrac{{16.17}}{2} = 136\).

Chọn B.