$N = \dfrac{\sqrt{x} + 3}{\sqrt{x} - 1}=\dfrac{\sqrt{x} - 1 + 4}{\sqrt{x} - 1} = 1 + \dfrac{4}{\sqrt{x} - 1} \, (ĐK: \, x \geq 0; \,x\ne 1)\\N \, nguyên \Leftrightarrow \dfrac{4}{\sqrt{x} - 1} \, nguyên\\\Leftrightarrow (\sqrt{x} - 1) \in Ư(4) = \left\{-4;-2;-1;1;2;4\right\}\\Do \, \sqrt{x} \geq 0 \Rightarrow \sqrt{x} - 1 \geq -1\\nên \, \sqrt{x} - 1 = \left\{-1;1;2;4\right\}\\\text{Ta có bảng giá trị:}$
$\begin{array}{|l|r}
\sqrt{x} - 1 & -1 &1 &2 &4\\
\hline
\sqrt{x} &0 & 2 &3 &5 \\
\hline
x & 0&4&9 &25&
\end{array}$
Vậy $S=0+4+9+25 = 38$
