, 2019

Solution:
Each time we draw in a diagonal, we create one new region, plus one new region for each intersection on that diagonal. So, the number of regions will be
$$1+ (\text{number of diagonals}) + (\text{number of intersections})$$
where (number of intersections) counts an intersection of three diagonals twice. Since no four diagonals can pass through a point, the only nonconstant term in our expression is the last one. To minimize this term, we want to maximize the number of triples of diagonals passing through the same point. Consider the set $S$ of triples of diagonals $A_{n} A_{n+3}$ that intersect at a single point. Each triple in $S$ must come from three consecutive diagonals, and two different triples can only have one diagonal in common, so $S$ has at $\operatorname{most}\left\lfloor\frac{2019}{2}\right\rfloor=1009$ triples. Hence the number of resulting pieces is at least
$$1 + 2019 + (2 \cdot 2019 - 1009) = 5049$$
To show that 5049 is attainable, we use the following construction. Let $B_{1} \ldots B_{1010}$ be a regular 1010-gon, and let $\ell_{n}$ denote the external angle bisector of $\angle B_{n-1} B_{n} B_{n+1}$. Let $A_{1}=\overleftrightarrow{B_{1009} B_{1010}} \cap \overleftrightarrow{B_{1} B_{2}}$, $A_{2018}=\overleftrightarrow{B_{1008} B_{1009}} \cap \overleftrightarrow{B_{1010} B_{1}}$, $A_{2019}=\ell_{1} \cap \ell_{1009}$, and for $n=1, \ldots, 1008$, define $A_{2n}=\ell_{n+1} \cap \overleftrightarrow{B_{n-1} B_{n}}$ and $A_{2n+1}=\ell_{n} \cap \overleftrightarrow{B_{n+1} B_{n+2}}$. It follows that, for all $n=0, \ldots, 1008$, $\overline{A_{2n-1} A_{2n+2}}$, $\overline{A_{2n} A_{2n+3}}$, and $\overline{A_{2n+1} A_{2n+4}}$ intersect at $B_{n+1}$.