20. Determine the value of the following expression:
$$
\left\lfloor\frac{11}{2010}\right\rfloor+\left\lfloor\frac{11 \times 2}{2010}\right\rfloor+\left\lfloor\frac{11 \times 3}{2010}\right\rfloor+\left\lfloor\frac{11 \times 4}{2010}\right\rfloor+\cdots+\left\lfloor\frac{11 \times 2009}{2010}\right\rfloor,
$$
where $\lfloor y\rfloor$ denotes the greatest integer less than or equal to $y$.
20. Answer: 10045
The number of grid points $(x, y)$ inside the rectangle with four corner vertices $(0,0),(11,0)$, $(0,2010)$ and $(11,2010)$ is
$$
(11-1) \times(2010-1)=20090 \text {. }
$$

There are no grid points on the diagonal between $(0,0)$ and $(11,2010)$ excluding these two points, since for any point $(x, y)$ on this segment between $(0,0)$ and $(11,2010)$, we have
$$
y=\frac{2010 x}{11} .
$$

But for an integer $x$ with $1 \leq x \leq 10, \frac{2010 x}{11}$ is not an integer.
Hence the number of grid points $(x, y)$ inside the triangle formed by corner points $(0,0)$, $(0,2010)$ and $(11,2010)$ is
$$
(11-1) \times(2010-1) / 2=20090 / 2=10045 .
$$

For any grid point $(x, y)$ inside the triangle formed by comer points $(0,0),(0,2010)$ and $(11,2010)$, we have
$$
1 \leq y \leq 2009, \quad 1 \leq x<\frac{11 y}{2010} .
$$

Thus, for any fixed positive integer $y$, there are the number of grid points satisfying the condition that $1 \leq x<\frac{11 y}{2010}$ is
$$
\left\lfloor\frac{11 y}{2010}\right\rfloor,
$$
as $\frac{11 y}{2010}$ is not an integer when $1 \leq y \leq 2009$. Thus the number of grid points $(x, y)$ inside the triangle formed by corner points $(0,0),(0,2010)$ and $(11,2010)$ is
$$
\left\lfloor\frac{11}{2010}\right\rfloor+\left\lfloor\frac{11 \times 2}{2010}\right\rfloor+\left\lfloor\frac{11 \times 3}{2010}\right\rfloor+\left\lfloor\frac{11 \times 4}{2010}\right\rfloor+\cdots+\left\lfloor\frac{11 \times 2009}{2010}\right\rfloor .
$$

Therefore the answer is 10045 .