1. **Finding the largest element of $A$:**

The roots of the quadratic equation $x^2 + mx + 2020 = 0$ are positive integers. Let the roots be $p$ and $q$. By Vieta's formulas, we have:
$$p + q = -m \quad \text{and} \quad pq = 2020$$
Since $p$ and $q$ are positive integers, they must be factors of 2020. The factor pairs of 2020 are:
$$(1, 2020), (2, 1010), (4, 505), (5, 404), (10, 202), (20, 101), (101, 20), (202, 10), (404, 5), (505, 4), (1010, 2), (2020, 1)$$
We need to find the pair that gives the largest value of $m = -(p + q)$. Calculating $p + q$ for each pair:
$$\begin{aligned} &1 + 2020 = 2021, \\ &2 + 1010 = 1012, \\ &4 + 505 = 509, \\ &5 + 404 = 409, \\ &10 + 202 = 212, \\ &20 + 101 = 121. \end{aligned}$$
The smallest sum is $121$, so the largest $m$ is:
$$m = -121$$

2. **Finding the smallest element of $B$:**

The roots of the quadratic equation $x^2 + 2020x + n = 0$ are negative integers. Let the roots be $-r$ and $-s$. By Vieta's formulas, we have:
$$-r - s = 2020 \quad \text{and} \quad rs = n$$
Since $r$ and $s$ are positive integers, they must be factors of $n$. We need to find the pair that gives the smallest value of $n = rs$. The factor pairs of 2020 are the same as above. Calculating $rs$ for each pair:
$$\begin{aligned} &1 \cdot 2020 = 2020, \\ &2 \cdot 1010 = 2020, \\ &4 \cdot 505 = 2020, \\ &5 \cdot 404 = 2020, \\ &10 \cdot 202 = 2020, \\ &20 \cdot 101 = 2020. \end{aligned}$$
The smallest $n$ is $2020$.

3. **Calculating $a + b$:**

The largest element of $A$ is $a = -121$ and the smallest element of $B$ is $b = 2020$. Therefore:
$$a + b = -121 + 2020 = 1899$$

4. **Finding the sum of the digits of $a + b$:**

The sum of the digits of $1899$ is:
$$1 + 8 + 9 + 9 = 27$$

The final answer is $\boxed{27}$