Pick a sequence of index $1=n_0<n_1<\ldots<n_k=2020$, and choose points $(x_{n_{i+1}},y_{n_i})$, $i=0,1,\ldots,k-1$. Clearly the whole polygon $C$ is blackened. We hope $\sum_{i=0}^{k-1} x_{n_{i+1}} y_{n_i} \le 4 |C|=2 \sum_{j=1}^{2019} (x_{j+1}y_j-x_jy_{j+1})$. Then I remember this problem, I guess choosing the sequence randomly and computing the expectation would work, but I'm not sure.

Consider the same problem for $n$ points. Notice that by affine transformation, the $y_0$ and $x_n$ can be arbitrary. Now, permit the x and y to be weakly increasing and decreasing respectively, closing the domain space. Consider the maximal value of minimum covering area / actual area. Notice that by multilinearity, every rectangle's x and y coordinates must be at a point's x and y coordinates. for each point involved, we can move the point left and down to decrease the area while preserving the minimal covering area without a rectangle passing through the point until either the minimal covering area with a rectangle passing through the point is equal to the minimal covering area without a rectangle passing through the point, or the point has reached the border, in other words, $x_i = x_{i-1}$ and $y_i = y_{i+1}$. Assume some point falls into the second case, we call the configuration degenerate. We also weakly induct on n that the minimum covering area / actual area is less than the desired maximum. The base cases are trivial, being equal to 2. Claim: Degenerate cases are either not maximal or have a minimum covering area / actual area less than the maximal one. Proof: If $(x_1, y_1)$ or $(x_{n-1}, y_{n-1})$ is degenerate, then we can delete $(x_0, y_0)$ or $(x_n, y_n)$ respectively to invoke the $n - 1$ inductive case. Otherwise, there is a sequence of consecutive degenerate points bounded on both sides by non-degenerate points. Then, let the non-degenerate point preceding the string of degenerate points be $(x_i, y_i)$ and the first degenerate point be $(x_{i+1} = x_i, y_{i+1})$. Then, by definition, a minimal covering configuration has a rectangle passing through $(x_i, y_i)$ and therefore $(x_{i+1}, y_{i+1})$, so let this rectangle also pass through $(x_j, y_j)$ where $j < i$, then since $y_jx_i+y_ix_{i+1} > y_jx_i = y_jx_{i+1}$, moving $(x_i, y_i)$ left infinitesimally preserves the minimum covering area, which does not pass through the moved $(x_i, y_i)$ point by the inequality, and decreases the area, implying that the degenerate case is not maximal. Therefore no point in any maximal configuration whose minimum covering area to actual ratio has not appeared for smaller values of n can be degenerate. Specifically, this means that the aforementioned process will result in every single point having a rectangle passing through it in any maximal state, meaning that since the minimum covering area cannot be reduced through a change in choice of rectangles, we have that $x_{i+1}y_{i+1} \leq (x_{i+2}-x_{i+1})(y_i-y_{i+1})$ so $\frac{x_{i+1}}{x_{i+2}} + \frac{y_{i+1}}{y_i} \leq 1$ for all $1 \leq i \leq n-1$. Now, rearrange the desired area ratio instead as $$\frac{2}{1-\frac{\sum_{i = 1}^{n-2} x_iy_{i+1}}{\sum_{i=0}^{n-1} x_{i+1}y_i}}$$ Now, since: $$\left(\frac{1}{\frac{x_{i+1}}{x_{i+2}} \cdot \frac{x_{i+2}}{x_{i+3}}} + \frac{1}{\frac{y_{i+1}}{y_{i}} \cdot \frac{y_{i+2}}{y_{i+1}}}\right) \cdot x_{i+1}y_{i+2} = x_{i+3}y_{i+2}+x_{i+1}y_{i}$$ And $\frac{1}{ab} + \frac{1}{(1-a)(1-b)} \geq 8$ so summing entirely, $\sum_{i = 1}^{n-2} x_iy_{i+1} \leq \frac{1}{4} \sum_{i=0}^{n-1} x_{i+1}y_i$, so $\frac{\sum_{i = 1}^{n-2} x_iy_{i+1}}{\sum_{i=0}^{n-1} x_{i+1}y_i} \leq \frac{1}{4}$, so the entire ratio is at most $\frac{8}{3}$. This is in fact sharp with the near equality case $(0, 1), (\frac{1}{2^n}, \frac{1}{2}), (\frac{1}{2^{n-1}}, \frac{1}{2^2}), \ldots, (\frac{1}{2}, \frac{1}{2^n}), (1, 0)$ approaching the same value from below, so we are done.