Shapovalov A.V.

For which $N$ can the numbers from 1 to $N$ be rearranged in such a way that the arithmetic mean of any group of two or more consecutive numbers is not an integer?
Odd $N$ do not work: in this case, the arithmetic mean $1 / 2(N+1)$ of all numbers is an integer.

For even $N=2 m$, the desired arrangement is, for example, as follows: $2,1,4,3,6,5, \ldots, 2 m, 2 m-1$.

Indeed, a group of $2 k$ consecutive numbers can be divided into pairs of numbers symmetrically located relative to the center of the group (for example, the group 1, 4, 3, 6, 5, 8 is divided into pairs (1, 8), (4, 5), (3, 6)). In each pair, the arithmetic mean is the same half-integer (the pair contains numbers of different parity). Therefore, the arithmetic mean of the entire group is also the same non-integer number.

A group of $2 k+1$ consecutive numbers consists of $2 k$ consecutive numbers and one number at the edge. If this number is swapped with its neighbor in the arrangement that is not part of the group, we get $2 k+1$ consecutive numbers, whose arithmetic mean is an integer. Therefore, the arithmetic mean of the numbers in the group differs from this integer by $1 / 2 k+1$.

## Answer

For all even $N$.

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