jmc

The sum of the first 20 positive even integers is $2+4+\cdots +40=2(1+2+\cdots +20)$. For all $n$, $1+2+\cdots +n=n(n+1)/2$, so $2(1+2+\cdots +20)=20\cdot 21=420$.

Let the four consecutive even integers be $n-6$, $n-4$, $n-2$, and $n$. Their sum is $4n-12=420$, so $n=\boxed{108}$.