8. (40 points) On a strip, numbers from 1 to 1598 were written in order and the strip was cut into several parts. It turned out that the arithmetic mean of all numbers in the first part is some natural number $n$, the second part - the number $2 n$, the third part - the number $3 n$, and so on. Explain for which $n$ this is possible.
Solution. Consider several consecutive natural numbers. If their quantity is even, then they can be paired with equal odd sums, hence their arithmetic mean is not an integer. If their quantity is odd, then their arithmetic mean is the middle number in this set. Therefore, all segments of the strip consist of an odd number of numbers.

The arithmetic mean of the numbers from 1 to $k$ is a natural number $n$, hence $k=2 n-1$. Thus, the first strip consists of numbers from 1 to $2 n-1$, the second strip consists of the single number $2 n$, the first number of the third strip is $2 n+1$, and the last number is $4 n-1$ (the sum of the extreme numbers must equal $6 n$), then the next number is $4 n$, followed by numbers from $4 n+1$ to $6 n-1$, and so on.

Thus, the last numbers of the strips alternate: $2 n-1, 2 n, 4 n-1, 4 n, 6 n-1, 6 n, \ldots, 2 k n-1, 2 k n$. Since the number 1598 is even and is the last number in some strip, it equals some $2 k n$. Hence $k n=799=17 \cdot 47$, so $n$ can be one of the divisors of this number: $1, 17, 47, 799$. It is clear from the solution that corresponding examples for each such $n$ can be found.