17. Find the smallest positive integer $n$, such that among any $n$ different positive integers, there must exist 8 different positive integers, the sum of which is divisible by 8.
17. Let set $A=\{1,9,17,25,33,41,49,8,16,24,32,40,48,56\}$, which contains 14 numbers, 7 of which leave a remainder of 1 when divided by 8, and the other 7 are divisible by 8. If we randomly select 8 numbers from $A$, assume that $t$ of these numbers leave a remainder of 1 when divided by 8 $(1 \leqslant t \leqslant 7)$, and the remaining $8-t$ numbers are multiples of 8, then the sum of the remainders of these 8 numbers when divided by 8 is $t (1 \leqslant t \leqslant 7)$. Therefore, the sum of any 8 numbers in $A$ is not divisible by 8, so the smallest positive integer $n \geqslant 15$. 

Furthermore, for any $n \geqslant 15$ positive integers, we can pair the odd numbers and the even numbers, forming at least 7 pairs (if $n$ is odd, one number remains unpaired; if $n$ is even, no number remains unpaired). The sum of each pair is even, so the sum of each pair when divided by 8 can only be $0, 2, 4, 6$. (1) If the 7 pairs belong to only 2 categories, then there must be 4 pairs in the same category, and the sum of these 8 numbers is divisible by 8. (2) If the 7 pairs belong to 4 different categories, with at least one pair in each category, then there must be at least 2 pairs in one category. By $0+0+2+6=8, 0+2+2+4=8, 2+4+4+6=16, 0+4+6+6=16$, we know that there must be 4 pairs whose sum of 8 numbers is divisible by 8. (3) If the 7 pairs belong to 3 categories, with at least one pair in each category, there are three scenarios: (a) If one category has at least 4 pairs, then the sum of these 8 numbers is divisible by 8. (b) If two categories each have 3 pairs, and the other category has one pair. (i) If there is no category with a remainder of 0, then by $2+2+6+6=16$ or $2+4+4+6=16$, we know that there must be 4 pairs whose sum of 8 numbers is divisible by 8. (ii) If there is no category with a remainder of 2, then by $0+4+6+6=16$ or $0+0+4+4=8$, we know that there must be 4 pairs whose sum of 8 numbers is divisible by 8. (iii) If there is no category with a remainder of 4, then by $2+2+6+6=16$ or $0+0+2+6=8$, we know that there must be 4 pairs whose sum of 8 numbers is divisible by 8. (iv) If there is no category with a remainder of 6, then by $0+2+2+4=8$ or $0+0+4+4=8$, we know that there must be 4 pairs whose sum of 8 numbers is divisible by 8. (c) If each category has at least 2 pairs, then by $2+2+6+6=16$ or $0+0+4+4=8$, we know that there must be 4 pairs whose sum of 8 numbers is a multiple of 8. 

In summary, when $n \geqslant 15$, there must be 8 numbers among the $n$ positive integers whose sum is a multiple of 8. Therefore, the smallest positive integer $n=15$.