We claim that if both players play optimally, then Smilla wins $\boxed{1021615}$ grams of gold. Note that if Leo keeps selecting the lighter chest (if both chests have equal mass, he chooses either one), then the heavier chest will always be at most $2020$ grams heavier then the lighter chest. This means that the heavier chest can have at most $\frac{1}{2} \cdot (1 + 2 + \dots + 2020) + 1010 = 1021615$ grams of gold. Now we show that Smilla can make the heavier chest at least $2020$ grams heavier than the lighter chest. At the start, Smilla sets aside the $2020$ gram nugget. If Leo chooses the red chest, then Smilla chooses a nugget with mass $0$ or $3 \pmod{4}$. If all have been used, Smilla would finally choose the $2020$ gram nugget. If Leo chooses the blue chest, same as for the red chest, except $0$ and $3 \pmod{4}$ is replaced with $1$ and $2 \pmod{4}$. Since the sum of all positive integers $0$ or $3 \pmod{4}$ less than $2020$ is equal to the sum of all positive integers $1$ or $2 \pmod{4}$ less than $2020$, the heavier chest is at least $2020$ grams heavier than the lighter chest, giving Smilla $\frac{1}{2} \cdot (1 + 2 + \dots + 2020) + 1010 = 1021615$ grams of gold.