, 2019

Solution:

In year $k$, the number of squirrels is
$$2(2(\cdots(2 \cdot 1 + 2019) + 2019) + \cdots) + 2019 = 2^{k} + 2019 \cdot \left(2^{k-1} + 2^{k-2} + \cdots + 1\right) = 2020 \cdot 2^{k} - 2019$$
and the number of rabbits is
$$4(4(\cdots(4 \cdot 1 - 2) - 2) - \cdots) - 2 = 4^{k} - 2 \cdot \left(4^{k-1} + 4^{k-2} + \cdots + 1\right) = \frac{4^{k} + 2}{3}$$
For the number of rabbits to exceed that of squirrels, we need
$$4^{k} + 2 > 6060 \cdot 2^{k} - 6057 \Leftrightarrow 2^{k} > 6059$$
Since $2^{13} > 6059 > 2^{12}$, $k = 13$ is the first year for which there are more rabbits than squirrels.