We observe that $2020 = 2^2 \cdot 5 \cdot 101$. Whenever $a + 2b = 2020$, $d = \gcd(a, b)$ must divide $2020$. So, to find the largest possible value of $d$, we have to look at the large divisors of $2020$. $d = 2020$ is not possible, since in this case $a + 2b \ge 2020 + 2 \cdot 2020 > 2020$. The next largest divisor is $1010$. Since $3 \cdot 1010 > 2020$, $d = 1010$ is also not possible. Next is $505$, and here we find that $a = 1010$ and $b = 505$ work, since $\gcd(1010, 505) = 505$, and $1010 + 2 \cdot 505 = 2020$. It follows that the largest possible value for $\gcd(a, b)$ is $505$. This solves part (a).

To solve part (b), we first consider two cases for a pair $(a, b)$ such that $a + 2b = 2020$:

(i) $a \mid b$: Let $b = k a$ for some integer $k \ge 1$. Then $a + 2k a = 2020$, i.e., $a = \frac{2020}{1 + 2k}$, and we have $\operatorname{lcm}(a, b) = k a = \frac{2020 k}{1 + 2k} = \frac{2020}{2 + 1/k}$. Hence $k$ should be minimized in order to minimize $\operatorname{lcm}(a, b)$. Since $1 + 2k \mid 2020$, we must take $k = 2$ (note that $3 \nmid 2020$). It follows that $a + 4a = 5a = 2020$, giving $a = 404$ and $b = 2a = 808$, so that the smallest least common multiple of $a$ and $b$ in this case is given by $\operatorname{lcm}(404, 808) = 808$.

(ii) $b \mid a$: Let $a = k b$ for some integer $k \ge 1$. Then $k b + 2b = 2020$, i.e., $b = \frac{2020}{2 + k}$, and we have $\operatorname{lcm}(a, b) = k b = \frac{2020 k}{2 + k} = \frac{2020}{1 + 2/k}$. Hence $k$ should be minimized in order to minimize $\operatorname{lcm}(a, b)$. Since $2 + k \mid 2020$, we must take $k = 2$ (again, $3 \nmid 2020$). It follows that $2b + 2b = 4b = 2020$, giving $b = 505$ and $a = 2b = 1010$, so that the smallest least common multiple of $a$ and $b$ in this case is given by $\operatorname{lcm}(1010, 505) = 1010$.

We now show that we may assume that either (i) or (ii) holds in order to solve the problem: Among all possible greatest common divisors of $a$ and $b$, where $a + 2b = 2020$, let us fix one of them, say $d$. Let $(a, b)$ be any pair such that $a + 2b = 2020$ with $\gcd(a, b) = d$. Put $a = \hat{a} d$ and $b = \hat{b} d$. Then $\hat{a}$ and $\hat{b}$ are relatively prime, so that $\operatorname{lcm}(\hat{a}, \hat{b}) = \hat{a} \hat{b} > \max\{\hat{a}, \hat{b}\}$. The smallest possible value for $\operatorname{lcm}(\hat{a}, \hat{b})$ (and hence also for $\operatorname{lcm}(a, b)$) is thus obtained if either $\hat{a} = 1$ or $\hat{b} = 1$ (in which case $\operatorname{lcm}(\hat{a}, \hat{b})$ is equal to either $\hat{b}$ or $\hat{a}$). But $\hat{a} = 1$ implies $a \mid b$ and $\hat{b} = 1$ implies $b \mid a$. Since we can repeat this argument for any possible value of $d$, we may assume that (i) or (ii) holds in order to solve the problem.

We conclude that the smallest possible least common multiple of $a$ and $b$, where $a + 2b = 2020$, is $\operatorname{lcm}(404, 808) = 808$.