The floor is a function 1

Let $y$ be a real number. Since $\lfloor y \rfloor$ is the integer less than or equal to $y$ , then $y - 1 < \lfloor y \rfloor \leq y$ .

In the same manner, $\frac{x}{20} + \frac{x}{17} - 2 < \left \lfloor \frac{x}{20} \right \rfloor + \left \lfloor \frac{x}{17} \right \rfloor \leq \frac{x}{20} + \frac{x}{17}$

$\frac{x}{20} + \frac{x}{17} - 2 < 2017 \leq \frac{x}{20} + \frac{x}{17}$

$\frac{37x}{340} - 2 < 2017 \leq \frac{37x}{340}$

$18534.59 \approx 2017 \times \frac{340}{37} \leq x < 2019 \times \frac{340}{37} \approx 18552.97$

$18535 \leq x \leq 18552$

Since $\left \lfloor \frac{18535}{20} \right \rfloor + \left \lfloor \frac{18535}{17} \right \rfloor = 926 + 1090 = 2016$ , we need to look at the next two integers which is a multiple of $20$ or $17$ which is $18540$ and $18547$ . Therefore the integral values of x that we need are $18540$ , $18541$ , $\ldots$ , $18546$ .

$18540 + 18541 + \ldots + 18546 = \boxed{129801}$

For start, let us take the equation as simple fraction $\dfrac{x}{20} + \dfrac{x}{17} = 2017$

upon solving we get $x \approx 18,534.595$

if we put this value in original equation we get $\left \lfloor \dfrac{18,534.595}{20} \right \rfloor + \left \lfloor \dfrac{18,534.595}{17} \right \rfloor = 2016$

which is one less than what is required. So, for increasing this value by one, we need a nearest number which is multiple of either 20 or 17, which in this case is 18,540. Now all values from 18,540 to 18,546 will satisfy the given equation as 18,547 is a multiple of 17, thereby will increase the value to 2018.

Finally sum of all values $= 18,540 + 18,541 +\cdots + 18,546 = \boxed{129801}$