Four. (50 points) Given the line segment $x+y=1(x \geqslant 0$, $y \geqslant 0)$ with 2020 points. Find the smallest positive integer $k$, such that for any 2020 points, there always exists a way to divide these 2020 points into two groups, where the sum of the y-coordinates of one group is no greater than $k$, and the sum of the x-coordinates of the other group is no greater than $k$ (these 2020 points can overlap).
Four, $k$ is at least 506.
Let these 2020 points be $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots, P_{2020}\left(x_{2020}, y_{2020}\right)$, satisfying $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{2020}$.
Let $M \cup N=\left\{P_{1}, P_{2}, \cdots, P_{2000}\right\}, M \cap N=\varnothing$.
Assume $M$ is the set for calculating the sum of the x-coordinates, and $N$ is the set for calculating the sum of the y-coordinates.
Take these 2020 points to be $\left(\frac{1010}{2021}, \frac{1011}{2021}\right)$.
If $k \leqslant 505$, then $|M| \leqslant 1010,|N| \leqslant 1009$.
Thus, $|M|+|N| \leqslant 2019506$.
Therefore, take $M=\left\{P_{1}, P_{2}, \cdots, P_{t-1}\right\}$,
$N=\left\{P_{t}, P_{t+1}, \cdots, P_{2020}\right\}$.
Hence, $\sum_{i=t}^{2020} y_{i}=\sum_{i=t}^{2020}\left(1-x_{i}\right)=2021-t-\sum_{i=t}^{2020} x_{i}$
$\leqslant 2021-t-(2021-t) x_{t}$
$\leqslant 2021-t-\frac{2021-t}{t} \sum_{i=1}^{t} x_{i}$
$<2021-t-\frac{2021-t}{t} \times 506$
$=2021-t-\frac{2021 \times 506}{t}+506$
$=506+2021-\left(t+\frac{2021 \times 506}{t}\right)$
$\leqslant 506+2021-2 \sqrt{2021 \times 506}$
$<506$.
Therefore, such $M$ and $N$ satisfy the conditions.
In conclusion, the minimum value of $k$ is 506.