5. Let's call the arrangement of $N$ grasshoppers on a line at various points of it $k$-successful if, after making the necessary number of moves according to the rules of leapfrog, the grasshoppers can achieve that the sum of pairwise distances between them decreases by at least $k$ times. For which $N \geqslant 2$ does there exist an arrangement that is $k$-successful for all natural $k$? (In leapfrog, on a move, one of the grasshoppers jumps to the point symmetric to it relative to another grasshopper.)

M. Svyatlovsky
Answer: for $N \geqslant 3$.

First solution. For $N=2$, no matter how the grasshoppers jump, the distance between them does not change.

Let $N \geqslant 3$. Place the 1st, 2nd, and 3rd grasshoppers on a line at points with coordinates $0,1, \sqrt{2}$, and call these grasshoppers $V, Q, R$. The others are seated arbitrarily at other distinct points. We will show that for any $k$, the grasshoppers can jump in such a way that the sum of pairwise distances decreases by at least a factor of $k$ (denote the initial sum by $P$).

Lemma. Suppose three grasshoppers sit on a line at distinct points, and the ratio of the distances from one of them to the other two is irrational. Then, no matter how many jumps they make over each other, they will always remain at distinct points, and the ratio of the distances from one of them to the other two will remain irrational.

Proof of the lemma. We will show that these conditions are preserved during a jump. Suppose, for some grasshopper, the ratio of the distances to the other two is rational, then these distances are of the form $a$ and $a q$, where $q$ is rational. Then the distance between the other two grasshoppers is non-zero and has the form $|a \pm a q|$, so the ratio of any two distances between the grasshoppers is rational, which is a contradiction. Suppose some grasshopper jumps over grasshopper $A$, then the distances from $A$ to both grasshoppers do not change, and thus the ratio of these distances remains irrational, in particular, the distances are different, and therefore the grasshoppers remain in distinct points.

Now, let's move on to the problem. Suppose the first few moves are made only by grasshoppers $V, Q, R$ and only over each other, according to the lemma, they will always remain in distinct points. We will show that after any number of such moves, they can further jump in such a way that the current minimum of pairwise distances between them decreases by at least a factor of two.

Suppose, without loss of generality, that at the current moment the minimum distance is between grasshoppers $V, Q$ with coordinates $a$ and $b$, and $R$ has coordinate $c$. Mark on the line all points with coordinates differing from $a$ by a multiple of $(a-b)$. Clearly, by jumping over each other, grasshoppers $V, Q$ can occupy any two adjacent marked points, then $R$ is not in a marked point (by the lemma, jumping only over each other, $V, Q, R$ remain in distinct points), then $V, Q$ can occupy two adjacent marked points between which $c$ lies, and the distance from $R$ to one of the grasshoppers will be no more than $|a-b| / 2$, that is, the smallest distance has decreased by at least a factor of 2.

Then, in several moves, grasshoppers $V, Q, R$ can decrease the smallest distance between them by at least a factor of 2, then in several moves again by at least a factor of 2, and so on, they can, in several moves, achieve that the distance between some two of them is some number $t$ less than $P /(2 N(N-1) \cdot k)$ - let them do just that. Call any two grasshoppers, between which the distance is $t$, good, and one of them call $D$, further these grasshoppers do not jump.

Next, any grasshopper not from the pair of good ones can, by jumping over the pair of good ones (in the appropriate order), move by $2 t$ in any direction on the line, and then can, in several jumps over them, end up at a distance less than $2 t$ from $D$, let all grasshoppers except the good ones make jumps in this way. Then the distance from any grasshopper to $D$ will be less than $2 t$, and thus all pairwise distances will be less than $4 t$, and thus their sum will be less than $4 t \cdot N(N-1) / 22$ we will explicitly provide the initial arrangement, which is $k$-successful for any natural number $k$.

First, for a given $N>2$, construct a finite geometric progression $1, q, \ldots, q^{N-1}$, with common ratio $q \in(0,1)$, for which the condition

$$
1=q+q^{2}+\ldots+q^{N-1}
$$

is satisfied. The required set exists for any integer $N>2$, since the equation

$$
q+q^{2}+\ldots+q^{N-1}-1=0
$$

has a solution on the interval $(0,1)$, as the left side changes sign at its ends.

Now place $N$ grasshoppers at the following initial points:

$$
0,1,(1+q),\left(1+q+q^{2}\right), \ldots,\left(1+q+\ldots+q^{N-3}\right),\left(1+q+\ldots+q^{N-2}\right)
$$

Consider the jump of the first grasshopper over the second; then its new coordinate will be $2=1+1=1+q+q^{2}+\ldots+q^{N-1}$. It turns out that the jumping grasshopper has become the rightmost, and all the grasshoppers are now located at the points

$$
1,(1+q),\left(1+q+q^{2}\right), \ldots,\left(1+q+\ldots+q^{N-2}\right),\left(1+q+\ldots+q^{N-1}\right)
$$

By shifting the origin of coordinates to the right by 1, we get the coordinates of the grasshoppers

$$
0, q,\left(q+q^{2}\right), \ldots,\left(q+\ldots+q^{N-2}\right),\left(q+\ldots+q^{N-1}\right)
$$

Thus, the grasshoppers have reduced their coordinates by exactly a factor of $q$. If the specified step (jump of the leftmost grasshopper over the nearest neighbor) is repeated $r$ times, then the pairwise distances will change by a factor of $q^{r}$, which will allow achieving any required reduction $1 / K$.