If either $a$ or $b$ is larger than 2020, then both must be for the product to be positive. However, the resulting product would be less than 1, so this case is impossible. Now, we see that $\left(\frac{2020-a}{a}, \frac{2020-b}{b}\right)$ must be in the form $\left(\frac{x}{y}, \frac{2 y}{x}\right)$, in some order, for relatively prime positive integers $x$ and $y$. Then $\frac{2020}{a}=\frac{x+y}{y}$ and $\frac{2020}{b}=\frac{x+2 y}{x}$, so $x+y$ and $x+2 y$ are relatively prime factors of 2020. Since $x+y<x+2 y<2(x+y)$, the only possibility is $x+y=4, x+2 y=5$. Thus, $(x, y)=(3,1)$, and $\left(\frac{2020-a}{a}, \frac{2020-b}{b}\right)=\left(3, \frac{2}{3}\right)$ because $a<b$. Solving gives $(a, b)=(505,1212)$.