Số nguyên nhỏ nhất thỏa mãn bất phương trình \(\dfrac{{2017 - x}}{{15}} + \dfrac{{2018 - x}}{{16}} + \dfrac{{17 + x}}{{2019}} + \dfrac{{18 + x}}{{2020}} \le 4\) là:

Ta có:

\(\begin{array}{l}\dfrac{{2017 - x}}{{15}} + \dfrac{{2018 - x}}{{16}} + \dfrac{{17 + x}}{{2019}} + \dfrac{{18 + x}}{{2020}} \le 4\\\dfrac{{2017 - x}}{{15}} + \dfrac{{2018 - x}}{{16}} + \dfrac{{17 + x}}{{2019}} + \dfrac{{18 + x}}{{2020}} - 4 \le 0\\ \Leftrightarrow \dfrac{{2017 - x}}{{15}} - 1 + \dfrac{{2018 - x}}{{16}} - 1 + \dfrac{{17 + x}}{{2019}} - 1 + \dfrac{{18 + x}}{{2020}} - 1 \le 0\\ \Leftrightarrow \dfrac{{2002 - x}}{{15}} + \dfrac{{2002 - x}}{{16}} + \dfrac{{x - 2002}}{{2019}} + \dfrac{{x - 2002}}{{2020}} \le 0\\ \Leftrightarrow  - \dfrac{{x - 2002}}{{15}} - \dfrac{{x - 2002}}{{16}} + \dfrac{{x - 2002}}{{2019}} + \dfrac{{x - 2002}}{{2020}} \le 0\\ \Leftrightarrow \left( {x - 2002} \right)\left( { - \dfrac{1}{{15}} - \dfrac{1}{{16}} + \dfrac{1}{{2019}} + \dfrac{1}{{2020}}} \right) \le 0\end{array}\)

Mà \(\dfrac{1}{{2019}} + \dfrac{1}{{2020}} - \dfrac{1}{{15}} - \dfrac{1}{{16}} < 0\) nên \(x - 2002 \ge 0 \Leftrightarrow {\rm{x}} \ge {\rm{2002}}\)

Vậy giá trị nguyên nhỏ nhất của \(x\) là \(2002\).