Let $m$ be a positive integer, and consider a $m\times m$ checkerboard consisting of unit squares. At the centre of some of these unit squares there is an ant. At time $0$, each ant starts moving with speed $1$ parallel to some edge of the checkerboard. When two ants moving in the opposite directions meet, they both turn $90^{\circ}$ clockwise and continue moving with speed $1$. When more than $2$ ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear.

Considering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard, or prove that such a moment does not necessarily exist.

[i]Proposed by Toomas Krips, Estonia[/i]
1. **Define the coordinate system and initial conditions:**
   - Place the checkerboard on the coordinate plane such that the center of the lower left square is \((1,1)\) and the center of the upper right square is \((m,m)\).
   - Note that collisions can only occur at \((x,y) \in [1,m]^2\).

2. **Define the function \(f(x,y)\):**
   - Let \(f(x,y)\) be the latest possible time at which an ant can be at \((x,y)\) moving straight up (considering the time right before a collision if it occurs).

3. **Determine the latest possible moment an ant falls off the board:**
   - A time \(\frac{1}{2}\) before exiting the board, an ant is at some \((x,y)\) moving in some direction. By a suitable rotation, we can assume WLOG that the direction is up.
   - The answer we seek is:
     \[
     \frac{1}{2} + \max_{(x,y) \in [1,m]^2} f(x,y).
     \]

4. **Main claim:**
   - We claim that:
     \[
     f(x,y) \leq y - 1 + \min\{x - 1, m - x\}.
     \]

5. **Proof of the claim:**
   - **Change the rules for ant movement:**
     - Post a collision, potentially swap ants so that an ant is always moving only up and right, or only down and left.
     - It is clear now that an ant moving up at \((x,y)\) had a path length at most \(x - 1 + y - 1\), so:
       \[
       f(x,y) \leq y - 1 + x - 1.
       \]
     - Now, change the rules so that an ant is always moving only up and left, or only down and right.
     - It is clear now that an ant moving up at \((x,y)\) had a path length at most \(m - x + y - 1\), so:
       \[
       f(x,y) \leq y - 1 + m - x.
       \]
     - Combining these, we get:
       \[
       f(x,y) \leq y - 1 + \min\{x - 1, m - x\}.
       \]
     - This proves the claim. \(\blacksquare\)

6. **Conclusion:**
   - This now implies that we can be sure that all ants have fallen off right after time:
     \[
     \frac{1}{2} + m - 1 + \frac{m - 1}{2} = \boxed{\frac{3}{2}m - 1}.
     \]
   - We may achieve equality by having an ant at \((1,1)\) start moving right, and an ant at \((m,1)\) start moving left.