4. Given that the sum of the squares of four positive integers is 2009, and the ratio of three of these numbers is $2: 3: 6$, then these four numbers are
4. $8,12,24,35$ or $10,15,30,28$ Detailed Explanation: Let $2009=(2 a)^{2}+(3 a)^{2}+(6 a)^{2}+b^{2}$, then $b^{2}=2009-49 a^{2}$ $=49\left(41-a^{2}\right)$, and we can set $b=7 c$, then $c^{2}=41-a^{2}$, solving this indeterminate equation we find $\left\{\begin{array}{l}a=4 \\ c=5\end{array}\right.$ or $\left\{\begin{array}{l}a=5 \\ c=4\end{array}\right.$, thus we know the four numbers are $8,12,24,35$ or $10,15,30,28$.