Sum of 8

Let the 8 consecutive positive integers be $n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6, n + 7$ .

Their sum is $8n + (1 + 2 + 3 + 4 + 5 + 6 + 7) = 8n + 28 = 4(2n + 7)$ , which is divisible by $4$ , but not by $8$ since $2n + 7$ is necessarily odd. Also, for $n = 2$ we have $2n + 7 = 11$ , a prime, and for $n = 3$ we have $2n + 7 = 13$ , a different prime, so for $n = 2,3$ the only common factor of $4(2n + 7)$ is $4$ . Thus the largest integer that will be guaranteed of dividing the sum of 8 positive consecutive integers is $\boxed{4}$ .

Note: In general, the largest integer that will be guaranteed of dividing the sum of $k$ positive integers will be $k$ if $k$ is odd and $\dfrac{k}{2}$ if $k$ is even.