Initially 255, set 2009 can be decomposed into the sum of squares of four positive integers, among which, the ratio of two numbers is $\frac{5}{14}$, and the ratio of the other two numbers is $\frac{1}{2}$. Write down this decomposition.
Solution: Let the required decomposition be
$$
2009=(14 x)^{2}+(5 x)^{2}+(2 y)^{2}+y^{2} \text {. }
$$

Obviously, $221 x^{2}+5 y^{2}=2009$, hence
$$
x^{2}=\frac{2009-5 y^{2}}{221}<10 \text {. }
$$

Thus, $x=1, 2$ or 3.
(1) If $x=1, y^{2}=\frac{2009-221 x^{2}}{5}=\frac{1788}{5}$,

no solution;
(2) If $x=2, y^{2}=\frac{2009-221 x^{2}}{5}=225$, hence $y=15$;
(3) If $x=3, y^{2}=\frac{2009-221 x^{2}}{5}=4$, hence $y=2$.
In summary, the required decompositions are
$$
\begin{array}{l}
2009=30^{2}+28^{2}+15^{2}+10^{2}, \\
2009=42^{2}+15^{2}+4^{2}+2^{2} .
\end{array}
$$
(Tian Yonghai, Suihua Educational Institute, Heilongjiang Province, 152054)