Problem 1011

We will show that the answer to the question is negative. Indeed, suppose that the required sequence exists, and let $a_{i}$ denote the position of the number $i$ in this sequence $(i=0,1,2, \ldots, 19)$. Then the enumeration $a_{0}, a_{1}, \ldots, a_{19}$ is a permutation of the first 20 positive integers, so

$$a_{0}+a_{1}+\ldots+a_{19}=1+2+\ldots+20=210$$

According to the condition, on the other hand,

$$\left|a_{0}-a_{10}\right|+\left|a_{1}-a_{11}\right|+\ldots+\left|a_{9}-a_{19}\right|=1+2+\ldots+10=55$$

If $P$ and $N$ denote the sum of the numbers with positive and negative signs, respectively, after removing the absolute value signs on the left side of (2), then from (1) we have $P+N=210$, and from (2) we get $P-N=55$. From this, $2P$ and $2N$ turn out to be odd, which is impossible since $P$ and $N$ are integers. This means that indeed no sequence with the stated property exists.

Remark. The problem is essentially a special case of the following: does there exist a sequence of length $2n$ in which the numbers $0,1, \ldots, n-1$ each appear exactly twice, and between any two occurrences of a number $k$, exactly $k$ numbers appear in the sequence? From the proof, it can be read that for the existence of such a sequence, it is necessary that the sum of the integers from $(n+1)$ to $(2n-1)$ be even, which means that $n$ must leave a remainder of 0 or 1 when divided by 4. It is not a simple question whether the found condition is sufficient.

In any case, such an arrangement exists for $n=1,4,5,8$ and 9: 00, 2132003, 0023421314, 2412174635003765, 141008473625328765.