jmc

The sum of the first 20 positive multiples of 5 is $5+10+15+\cdots +95+100=5(1+2+\cdots +20)$. For all $n$, $1+2+\cdots +n=n(n+1)/2$, so $5(1+2+\cdots +20)=5\cdot 20\cdot 21/2=1050$.

The sum of the first 20 positive even integers is $2+4+6+8+\cdots +38+40=2(1+2+\cdots +20)=2\cdot 20\cdot 21/2=420$. The difference is $1050-420=\boxed{630}.$

Alternatively, we can subtract the sums, to get $$\begin{array}{rl}5(1+2+\cdots +20)-2(1+2+\cdots +20)& =3(1+2+\cdots +20)\\ & =3\cdot \frac{20\cdot 21}{2}\\ & =630.\end{array}$$