We call a time point "special" if at the time, there some ant crawled up/down from the hole. It is easy to see that we just need to solve the problem in case $|T| = 45$ (if $|T| < 45$, we consider some more special time points and that will make the below argument stronger).

Consider the system of coordination Oxy in which Ox expresses to the location of holes on the line and Oy expresses to the time. The coordination of holes are $x_1, x_2, \dots, x_{44}$ and special times are $y_1, y_2, \dots, y_{45}$.

An ant move from point $(x_a, y_b)$ to point $(x_c, y_d)$ if this ant crawled from the hole $x_a$ at the time $y_b$ and move to the hole $x_c$ at the time $y_d$. Since that speed of ant did not change, the graph express for that moving is a segment connect two correspondent points.

Thus in total, we have $2017$ segments and since the speed of ants are pairwise distinct, the lines have different directions. To finish the problem, we need to show that at least two segment among them does not intersect. Note that the number of end points of these segment is no more than $45 \cdot 44 = 1980 < 2017$ then we will prove the generalization version of this problem as: If there are $n$ points on the plane, then there are no more than $n$ segments connect them such that no two segments parallel or overlap.

(*)

We shall prove (*) by induction. It is easy to check for $n = 2, 3$.

For $n \ge 4$, suppose that the argument is true for $n-1$ points. We consider two cases:

1. If among $n$ points, there is one point is the end point of at most one segment, then by removing it (also the segment has it as the end point), we come to the case of $n-1$ points.

2. If each point is the end point of at least two segments, then we will show that the number of segments is exactly the same as the number of points. Indeed, if there is some point is the end point of three segments, for example point $A$ is the end point of $AB, AC, AD$. If point $A$ lies inside triangle $BCD$ then the second segment from $B$ cannot cut both $AC$ and $AD$. Otherwise, we can suppose that the ray $AC$ lie between two rays $AB, AD$ then the second segment from $C$ cannot cut both $AB$ and $AD$. This implies that each point is the end point of exactly two segments then the number of segment is $n$.

Hence (*) also holds for $n$ which complete the proof. ■