10. At the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip is shifted one position clockwise relative to its initial position? (S. Berlov)
Answer: 50. Solution: Example: The chip 50 is sequentially exchanged 99 times with the next one counterclockwise.

Evaluation. We reason by contradiction. Let $k<50$. First proof. We will consider the shifts of the chips relative to their initial positions, with shifts clockwise counted as positive and counterclockwise as negative. Then, when two chips are exchanged, 1 is added to the shift of one and 1 is subtracted from the shift of the other. Suppose after several moves all chips have shifted one position clockwise. Then the total shift of the chip numbered $k$ is $100 t_{k} + 1$, where $t_{k}$ is the number of complete revolutions of this chip (clockwise revolutions are counted with a plus sign, and counterclockwise with a minus sign). Since $k<50$, chips numbered 1 and 51 could not have exchanged places, and therefore they must have made the same number of complete revolutions, i.e., $t_{1} = t_{51}$. Similarly, $t_{2} = t_{52}, \ldots, t_{50} = t_{100}$. Therefore, the sum of all shifts of all chips is $100(2 t_{1} + \ldots + 2 t_{50} + 1)$. This sum must be 0, as the sum of shifts is 0 for each move. But it is not 0, since the sum in parentheses is odd. Contradiction.

Second proof. At any moment, we consider the \{lit painted\} arc from chip 100 to chip 1 clockwise. Since chips 100 and 1 cannot be exchanged in one move, any specific chip $\mathrm{m}$ $(2 \leq m \leq 99)$ could only enter the painted arc or leave the painted arc by exchanging with one of the chips 1 or 100. Since initially and at the end, chip $m$ was not on the painted arc, it must have made an equal number of \{lit entries\} to the painted arc and \{lit exits\} from the painted arc. For $m \leq 50$, chip $m$ could not have exchanged with chip 100, so it could only make \{lit entries\} or \{lit exits\} by exchanging with chip 1. On \{lit entry\}, chip 1 makes a shift of 1 clockwise, and on \{lit exit\}, 一 1 counterclockwise. We conduct similar reasoning for chips $m \geq 51$, which cannot exchange with chip 1. Thus, we find that chips 1 and 100 will make the same number of shifts clockwise and counterclockwise, so they will remain in their positions. Contradiction.