Solution:

We denote the centers using the coordinates of the squares, so that the centers of the bottom row are the points $(0,0), \ldots, (2019,0)$ and those of the top row are $(0,1), \ldots, (2019,1)$.

a.
Consider the set of segments $S$ that connect the following pairs of points:
(i) $(0,0)-(k, 1)$ for $k=1, \ldots, 2019$;
(ii) $(0,0)-(0,1)$;
(iii) $(k, 0)-(0,1)$ for $k=1, \ldots, 2019$.
In the previous list no pair appears twice, hence the corresponding segments are also distinct. Since in the first point we have indicated $2019$ pairs, in the second $1$ pair, and in the third $2019$ pairs, in total those listed are $4039$ distinct pairs.
Let us choose any two segments $s_{1}, s_{2} \in S$. If neither is of type (iii), then $s_{1} \cap s_{2} = (0,0)$; similarly, if neither is of type (i), then $s_{1} \cap s_{2} = (0,1)$. If instead $s_{1}$ is of type (i) and $s_{2}$ is of type (iii), or vice versa, then $s_{1}$ and $s_{2}$ are the (internal) diagonals of the convex, non-crossed quadrilateral whose vertices are $(0,0), (x_{s_{2}}, 0), (x_{s_{1}}, 1), (0,1)$, so they intersect.
Summing up, we have the following two facts:
- $|S| = 4039$
- every pair of segments in $S$ intersects.
If we had at most $4038$ colors available, by the pigeonhole principle at least two segments in $S$ would have the same color, but this is incompatible with the requirement that non-disjoint segments have different colors.

b.
Let us enumerate the $4039$ colors in this way: $-2019, \ldots, -1, 0, 1, \ldots, 2019$.
A valid coloring of the segments is the following: the segment between $(x_{1}, 0)$ and $(x_{2}, 1)$ is colored with color $x_{2} - x_{1}$.
Let us show that any two segments $s_{1}, s_{2}$ of the same color $c$ do not intersect. Let $(x_{1}, 0)$ be the first vertex of $s_{1}$: then the second must be $(x_{1} + c, 1)$; similarly let $(x_{2}, 0)$ be the first vertex of $s_{2}$, and consequently $(x_{2} + c, 1)$ the second. Then $s_{1}$ and $s_{2}$ are two opposite sides of the quadrilateral having as vertices $(x_{1}, 0), (x_{2}, 0), (x_{2} + c, 1), (x_{1} + c, 1)$. This quadrilateral is a (non-crossed) parallelogram because the other two sides are parallel and have length $c$. Hence $s_{1}$ and $s_{2}$ do not intersect.