7.107 There are two grasshoppers at the two endpoints of the line segment $[0,1]$, and some points are marked within the line segment. Each grasshopper can jump over the marked points such that the positions before and after the jump are symmetric with respect to the marked point, and they must not jump out of the range of the line segment $[0,1]$. Each grasshopper can independently jump once or stay in place, which counts as one step. How many steps are needed at minimum to ensure that the two grasshoppers can always jump to the same small segment divided by the marked points on $[0,1]$?
[Solution] The segments into which the segment $[0,1]$ is divided by the marked points are called line segments. It is easy to prove that if 3 marked points $\frac{9}{23}, \frac{17}{23}$, and $\frac{19}{23}$ are taken, then after each of the two fleas jumps once, they cannot land in the same line segment (see the figure below). Therefore, the minimum number of steps required must be greater than 1.

Next, we prove that, regardless of how many marked points are taken and how they are placed on the segment $[0,1]$, it is always possible to make the two fleas jump to the longest line segment after two jumps (if there is more than one such line segment, any one of them can be chosen). By symmetry, it is only necessary to prove this statement for the flea starting at point 0.

Let the length of the chosen longest line segment be $s$, and its left endpoint be $\alpha$. If $\alpha < s$, then the flea can jump over the marked point $\alpha$ in one step and land in the chosen line segment $[\alpha, \alpha+s]$ (if $\alpha = 0$, the flea is already in the chosen line segment and does not need to jump at all). If $\alpha \geqslant s$, consider the interval $\left[\frac{\alpha-s}{2}, \frac{\alpha+s}{2}\right]$, which has a length of $s$. Therefore, this interval must contain at least one marked point, denoted as $\beta$. Otherwise, the length of the line segment containing this interval would be greater than $s$, which is impossible. Jumping over $\beta$ in one step, the flea will land at point $2\beta \in [\alpha-s, \alpha+s]$.

If $2\beta \in [\alpha, \alpha+s]$, then jumping over point $\alpha$ again, the flea will definitely land in the chosen line segment $[\alpha, \alpha+s]$.