3. Let $n \geqslant 2$ be a positive integer. Initially, there are $n$ grasshoppers on a straight line, and they are not all at the same point. For any given positive real number $\lambda$, the following "move" can be defined:
(1) Select any three grasshoppers, let them be at points $A$ and $B$, with $A$ to the left of $B$;
(2) Let the grasshopper at point $A$ jump to a point $C$ on the line to the right of $B$, such that $\frac{B C}{A B}=\lambda$.

Determine all possible positive real numbers $\lambda$, such that for any given point $M$ on the line and any initial positions of the $n$ grasshoppers, it is always possible to move all the grasshoppers to the right of $M$ after a finite number of moves.
To make the flea jump as far to the right as possible, a reasonable strategy is to choose the position of the leftmost flea as point $A$ and the position of the rightmost flea as point $B$ at each move. According to this strategy, assume that after $k$ moves, the maximum distance between the fleas is $d_{k}$, and the minimum distance between any two adjacent fleas is $\delta_{k}$. Clearly,
$$
d_{k} \geqslant(n-1) \delta_{k} \text {. }
$$

After the $(k+1)$-th move, a new distance between two adjacent fleas will be $\lambda d_{k}$. If this is the new minimum value, then $\delta_{k+1}=\lambda d_{k}$; if it is not the minimum value, then obviously $\delta_{k+1} \geqslant \delta_{k}$. In either case, we always have
$$
\frac{\delta_{k+1}}{\delta_{k}} \geqslant \min \left\{1, \frac{\lambda d_{k}}{\delta_{k}}\right\} \geqslant \min \{1,(n-1) \lambda\} .
$$

Therefore, as long as $\lambda \geqslant \frac{1}{n-1}$, we have $\delta_{k+1} \geqslant \delta_{k}$ for any $k$. This means that the minimum distance between any two adjacent fleas will not decrease. Hence, after each move, the position of the leftmost flea will move to the right by at least a positive constant. Eventually, all the fleas can jump to the right of any given point $M$.
The proof below: If $\lambda\omega_{k}$, then after the move, the flea that has jumped occupies the new rightmost position $w_{k+1}=c$, and by $b \leqslant w_{k}$ we get
$$
s_{k+1}-s_{k}=\frac{1+\lambda}{\lambda}(c-b) \geqslant \frac{1+\lambda}{\lambda}\left(w_{k+1}-w_{k}\right) .
$$

If $c \leqslant w_{k}$, then
$$
w_{k+1}-w_{k}=0, s_{k+1}-s_{k}=c-a>0 .
$$

Thus, the above inequality still holds.
Consider the following sequence
$$
z_{k}=\frac{1+\lambda}{\lambda} \cdot w_{k}-s_{k}, k=0,1,2, \cdots
$$

Then $z_{k+1}-z_{k} \leqslant 0$, meaning the sequence is non-increasing.
Therefore, for all $k$, we always have $z_{k} \leqslant z_{0}$.
Assume $\lambda n \lambda$. We can write $z_{k}$ as $z_{k}=(n+\mu) w_{k}-s_{k}$, where $\mu=\frac{1+\lambda}{\lambda}-n>0$.
Thus, we get the inequality
$$
z_{k}=\mu w_{k}+\left(n w_{k}-s_{k}\right) \geqslant \mu w_{k} \text {. }
$$

Hence, for all $k$, we always have $w_{k} \leqslant \frac{z_{0}}{\mu}$. This means the position of the rightmost flea will never exceed a constant, which depends on $n$, $\lambda$, and the initial positions of the fleas, but not on how they move. The final conclusion is: the possible values of $\lambda$ are all real numbers not less than $\frac{1}{n-1}$.