, 2020

Solution:
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 $\left(\frac{x}{y}, \frac{2y}{x}\right)$, in some order, for relatively prime positive integers $x$ and $y$.

Then $\frac{2020}{a} = \frac{x+y}{y}$ and $\frac{2020}{b} = \frac{x+2y}{x}$, so $x+y$ and $x+2y$ are relatively prime factors of $2020$.

Since $x+y < x+2y < 2(x+y)$, the only possibility is $x+y = 4$, $x+2y = 5$. 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 $(a, b) = (505, 1212)$.

Solution:
We rearrange to find that $(a+2020)(b+2020) = 2 \cdot 2020^{2}$. Note that $a+2020$ and $b+2020$ are both less than $4040 < 101^{2}$, so they must both be divisible by $101$. Hence, we divide out a factor of $101^{2}$ and solve the equivalent problem of $(a' + 20)(b' + 20) = 800$, where $a' = \frac{a}{101}$ and $b' = \frac{b}{101}$.

Because each factor must be less than $\frac{4040}{101} = 40$, we get that $(a' + 20, b' + 20) = (25, 32)$, which yields $(a, b) = (505, 1212)$.