Let $I_r={\int }_0^{r}x|\sin \pi x|dx$. Apply the king's reflection on $[0,r]$, $x\to r-x$:

$I_r={\int }_0^r(r-x)|\sin \pi (r-x)|dx={\int }_0^r(r-x)|\sin \pi x|dx$ (since $|\sin \pi (r-x)|=|\sin \pi x|$ for integer $r$).

Adding the two expressions: $2I_r=r{\int }_0^r|\sin \pi x|dx$, so $I_r=\frac{r}{2}{\int }_0^r|\sin \pi x|dx$.

${\int }_0^1|\sin \pi x|dx=\frac{2}{\pi }$ (one full positive arch). Over $[0,r]$ with $r$ a positive integer, the integral is $\frac{2r}{\pi }$.

$\therefore I_r=\frac{r}{2}\cdot \frac{2r}{\pi }=\frac{r^2}{\pi }$.

Then $\pi I_r=r^2$ and $\sqrt{\pi I_r}=r$ (exact integer for integer $r$).

$\sum _{r=1}^{20}\lfloor r\rfloor =\sum _{r=1}^{20}r=\frac{20\cdot 21}{2}=210$.