Let $a_{1}, a_{2}, \ldots$ be a sequence of integers that has infinitely many positive and infinitely many negative elements, and for every $n$, it holds that $a_{1}, a_{2}, \ldots, a_{n}$ leave pairwise distinct remainders when divided by $n$. How many times does $2008$ appear in the sequence?
Solution. In the sequence, no integer can appear more than once: if for some $i<j$ we have $a_{i}=a_{j}$, then for $n=j$ the condition of the problem is not satisfied, since among the first $j$ elements, the $i$-th and the $j$-th (since they are equal) give the same remainder when divided by $j$. Similarly, we show that (for every $k$) the difference (in absolute value) between any two of the first $k$ elements of the sequence can be at most $k-1$. Indeed, suppose that among the numbers $a_{1}, a_{2}, \ldots, a_{k}$ there are two whose difference is $a_{j}-a_{i}=n \geq k$. Then among the first $n$ elements of the sequence, the $i$-th and the $j$-th (since their difference is exactly $n$) would give the same remainder when divided by $n$. These two properties together mean that (for every $k$) the set $\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ consists of $k$ consecutive integers. Furthermore, since the sequence contains infinitely many (different) positive integers, there is certainly one greater than 2008, and similarly, due to the negative elements, there is one less than 2008. If we choose $n$ so that among the first $n$ terms of the sequence there is one greater than 2008 and one less than 2008, then one of the first $n$ terms is 2008.

Therefore, 2008 appears exactly once in the sequence (and this can be said for any integer).

Remark. The condition of the problem is satisfied by infinitely many sequences: for any integer $t$, for example, the sequence $t, t+1, t-1$, $t+2, t-2, t+3, t-3, \ldots$ does.