Problem 1569

10. Proof: Let the first $2n$ natural numbers be written in any order as $a_{1}, a_{2}, a_{3}, \cdots, a_{2n-1}, a_{2n}$, where $1 \leqslant a_{i} \leqslant 2n$, $a_{i} \neq a_{j}(i \neq j),(i, j=1,2,3, \cdots, 2n)$, then the corresponding sum numbers are $1+a_{1}, 2+a_{2}, \cdots, 2n-1+a_{2n-1}$, $2n+a_{2n}$.

These sum numbers, when divided by $2n$, can only result in one of the following $2n$ remainders: $0,1,2, \cdots, 2n-2,2n-1$.

If we can prove that all $2n$ remainders cannot appear, the problem is solved. A direct approach is difficult, so we consider the opposite.

Suppose these $2n$ remainders all appear, then their sum is $r_{2n}=0+1+2+\cdots+(2n-1)=2n^2-n>0$.
This sum should be congruent modulo $2n$ to the sum of the $2n$ sum numbers divided by $2n$ (i.e., the remainders are the same), which indicates a remainder of zero. This is a contradiction.