Mumchov D:

On a circle of length 2013, 2013 points are marked, dividing it into equal arcs. A chip is placed at each marked point. We define the distance between two points as the length of the shorter arc between them. For what largest $n$ can the chips be rearranged so that there is again one chip at each marked point, and the distance between any two chips that were initially no more than $n$ apart has increased?
Evaluation. Let's number the points and the chips on them in a clockwise direction with consecutive non-negative integers from 0 to 2012. Consider an arbitrary permutation and the chips with numbers 0, 671, and 1342, initially located at the vertices of an equilateral triangle. The pairwise distances between them are 671. After the permutation, the sum of the pairwise distances between these chips will not exceed the length of the circle, which means the distance between some two of them will not exceed \(2013 / 3 = 671\); therefore, the distance between these two chips will not increase. Thus, for \(n \geq 671\), the required permutation is impossible.

Example of the desired permutation for \(n=670\). Each chip with a number \(i \leq 1006\) will be moved to the point with the number \(a_{i} = 2i\), and each chip with a number \(i \geq 1007\) will be moved to the point with the number \(a_{i} = 2i - 2013\). In other words, \(a_{i}\) is the remainder of the division of \(2i\) by 2013. It is easy to see that each point has received one chip. It remains to show that the distances between pairs of chips, initially separated by no more than 670, will increase under this permutation.

Consider arbitrary chips with numbers \(i\) and \(j\); let the distance between them be \(d \leq 670\). Then one of the arcs between points \(a_{i}\) and \(a_{j}\) will have a length of \(2d\), meaning the distance between these points is \(d' = \min \{2d, 2013 - 2d\}\). But note that \(2d > d\) and \(2013 - 2d > d\). Therefore, \(d' > d\), as required.

## Answer

\(n=670\).