Problem 184

Since the robot moves along the number line starting from the origin and follows a cycle of moving 4 steps forward and then 3 steps back:

1. In a full cycle of 7 seconds, the robot effectively moves 1 step forward.
2. Therefore, after $7n$ seconds (where $n$ is a natural number), the robot will be at position $n$ on the number line.

To find the positions of $x_{2007}$ and $x_{2011}$:

- Divide 2007 by 7 to find the full cycles and the remainder:
$$2007 = 7 \times 286 + 5$$
This tells us that after 2007 seconds, the robot completed 286 full cycles and an additional 5 seconds into the next cycle.

- By the pattern of the robot's movement, the position $x_{2007}$ will be:
- After 286 cycles, at position 286.
- Then moves forward 4 steps to 290.
- Then moves back 1 step to 289 (because the robot goes back 3 steps in total, but we only need to account for 1 step in the remaining 5 seconds).

Therefore, $x_{2007} = 289$.

- Similarly, divide 2011 by 7 to find the full cycles and the remainder:
$$2011 = 7 \times 287 + 2$$
This tells us that after 2011 seconds, the robot completed 287 full cycles and an additional 2 seconds into the next cycle.

- Following the movement pattern:
- After 287 cycles, at position 287.
- Then moves forward 2 steps to 289.

Hence, $x_{2011} = 289$.

So the final result is:
$$x_{2007} - x_{2011} = 289 - 289 = 0$$

We \boxed{0} as the answer for $x_{2007} - x_{2011}$.