8 、Arrange the n consecutive positive integers from 1 to $n(n>1)$ in a sequence such that the sum of every two adjacent terms is a perfect square. The minimum value of the positive integer $n$ is $\qquad$ .
Answer: 15.
For example, a sequence that can be generated is
$$
(8,1,15,10,6,3,13,12,4,5,11,14,2,7,9) \text {. }
$$

Solution: This is an operational problem, which would be cumbersome to express in words, so it is suitable as a fill-in-the-blank question for direct operation. Let the set of these $n$ consecutive positive integers be $M=\{1,2, \cdots, n\}$. Since $n>1$, $M$ must contain 2, and $2+7=9$, so $n \geq 7$. When $n=7$, the numbers from 1 to 7 can be paired into three segments that meet the conditions: $(1,3,6),(2,7),(4,5)$, but they cannot be connected into a 7-term sequence, so more numbers need to be added. Adding 8 can expand the first segment to $(8,1,3,6)$, and adding 9 can expand the second segment to $(2,7,9)$, but the new three segments still cannot be connected, so more numbers need to be added, i.e., $n \geq 10$. If the previous numbers are adjacent to $8,9,10$, only $8+1=9,9+7=16,10+6=16$ are possible. These segments expand to $(8,1,3,6,10),(2,7,9)$, $(4,5)$, but they still cannot be connected. The next square number 25 should be used. The smallest number from 1 to 10 that can pair to sum to 25 is 15, so $n \geq 15$. When $M=\{1,2, \cdots, 15\}$, the sequence can be arranged as follows:
$$
(8,1,15,10,6,3,13,12,4,5,11,14,2,7,9) \text {. }
$$