Let $n \geq 2$ be a positive integer. Initially, $n$ fleas sit on a horizontal line, not all at the same point.

For a positive real number $\lambda$, define a step as follows:

choose two fleas sitting at points $A$ and $B$, where $A$ is to the left of $B$;

the flea at $A$ jumps to a point $C$ on the line, which is to the right of $B$, such that $BC / AB = \lambda$. Determine all values of $\lambda$ for which it is true that, no matter how the point $M$ is chosen on the line, and no matter how the initial positions of the $n$ fleas are chosen, there exists a finite sequence of steps such that after performing these steps, all fleas are to the right of $M$.
For simplicity, let's call the fleas points. First, we show that the statement is always true for $\lambda \geq \frac{1}{n-1}$. We show that no matter how far the points can move, if we always jump the leftmost point over the rightmost one. In this case, a greedy algorithm will move the points beyond $M$. In the first $n-1$ steps, each point lands in a different position, which we can consider as the initial position for the rest. Let the distances between adjacent points be $d_{1}, d_{2}, \ldots, d_{n-1}$. After one jump, the new distances are: $d_{2}, d_{3}, \ldots, d_{n-1},\left(d_{1}+d_{2}+\cdots+d_{n-1}\right) \cdot \lambda$. Since

$$
\left(d_{1}+d_{2}+\cdots+d_{n-1}\right) \cdot \lambda \geq \frac{d_{1}+d_{2}+\cdots+d_{n-1}}{n-1} \geq \min \left(d_{1}, \ldots, d_{n-1}\right)
$$

the smallest distance remains unchanged. Let this distance be $d$. The rightmost point thus moves at least $d$ to the right in each step. Since $d>0$, after some time it can reach any distance. Then, in $n-1$ steps, the other points also move beyond $M$, which is what we wanted to prove.

Now we show that if $\lambda < \frac{1}{n-1}$, the statement is not always true. Let the positions of the points be $x_{1}, x_{2}, \ldots, x_{n}$, and let $s_{k} = x_{1} + x_{2} + \cdots + x_{k}$ and $w_{k} = x_{k} - x_{1}$. Then

$$
\frac{1+\lambda}{\lambda} \cdot w_{k} - s_{k} = \frac{1+\lambda}{\lambda} \cdot (x_{k} - x_{1}) - (x_{1} + x_{2} + \cdots + x_{k})
$$

Since

$$
\frac{1+\lambda}{\lambda} \cdot (x_{k} - x_{1}) - (x_{1} + x_{2} + \cdots + x_{k}) = \frac{1+\lambda}{\lambda} \cdot x_{k} - \left(\frac{1+\lambda}{\lambda} + 1 + 1 + \cdots + 1\right) \cdot x_{1} - (x_{2} + x_{3} + \cdots + x_{k-1})
$$

$$
= \frac{1+\lambda}{\lambda} \cdot x_{k} - \left(\frac{1+\lambda}{\lambda} + k - 1\right) \cdot x_{1} - (x_{2} + x_{3} + \cdots + x_{k-1})
$$

$$
= \frac{1+\lambda}{\lambda} \cdot x_{k} - \left(\frac{1+\lambda}{\lambda} + k - 1\right) \cdot x_{1} - (x_{2} + x_{3} + \cdots + x_{k-1})
$$

$$
= \frac{1+\lambda}{\lambda} \cdot x_{k} - \left(\frac{1+\lambda}{\lambda} + k - 1\right) \cdot x_{1} - (x_{2} + x_{3} + \cdots + x_{k-1})
$$

Since $\lambda < \frac{1}{n-1}$, we have $\frac{1+\lambda}{\lambda} - n > 0$. Let $\mu = \frac{1+\lambda}{\lambda} - n$. Then

$$
\frac{1+\lambda}{\lambda} \cdot w_{k} - s_{k} = \mu \cdot w_{k} + (n \cdot w_{k} - s_{k}) \geq \mu \cdot w_{k}
$$

Therefore, $w_{k} \leq \frac{\frac{1+\lambda}{\lambda} \cdot w_{1} - s_{1}}{\mu}$, so $w_{k}$ is bounded, and we are done.

Thus, the solution is: $\lambda \geq \frac{1}{n-1}$.