4. There are two fleas at the two endpoints of the line segment $[0, 1]$. Some points are marked within the line segment. Each flea can jump over the marked points such that the positions before and after the jump are symmetric about the marked point, and they must not jump out of the range of the segment $[0, 1]$. Each flea 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 fleas can always jump to the same small segment divided by the marked points on $[0, 1]$?
4. The segment [0, 1] is divided into smaller segments by certain fixed points, with lengths $\frac{17}{23}$ and $\frac{19}{23}$. Therefore, the two fleas cannot land on the same segment after each taking one step (see the following supplement). From this, we can conclude that the minimum number of steps required must be 1.

The following proof shows that, regardless of the number of fixed points and how they are placed on the segment $[0,1]$, it is always possible to make both fleas jump into the longest segment after two jumps (if there are multiple such segments, we can choose one of them). By symmetry, it is sufficient to prove this for the flea starting at point 0.

Let the length of the chosen longest segment be $s$, and $\alpha$ be one of its endpoints.

If $\alpha < s$ (see the following figure), the flea can jump over the fixed point $\alpha$ in one step and land in the selected segment $[\alpha, \alpha+s]$ (if $\alpha=0$, the flea is already in the selected 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 $s$ (see the following figure). This interval must contain at least one fixed point $\beta$. Otherwise, the length of the segment containing this interval would be greater than $s$, which is impossible. By jumping over $\beta$, the flea will land at point $2 \beta \in [\alpha-s, \alpha+s]$. If $2 \beta \notin [\alpha, \alpha+s]$, then by jumping over point $\alpha$ again, the flea will definitely land in the selected segment $[\alpha, \alpha+s]$.