jmc

The first 8 odd positive integers are 1, 3, $\dots$, 15. The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so their sum is $(1+15)/2\cdot 8=64$.

Let the 5 consecutive even integers be $a$, $a+2$, $a+4$, $a+6$, and $a+8$. Their sum is $5a+20$. But this is also $64-4=60$, so $5a+20=60$. Solving for $a$, we find $a=\boxed{8}$.