$31 \cdot 38 \sqrt{x}+\sqrt{y}=\sqrt{2009}$ The positive integer solutions $(x, y)$ of, the maximum value of $x+y$ is
(A) 1189.
(B) 1517.
(C) 1657.
(D) 1749.
(Chinese Shaanxi Province Xi'an Junior High School Mathematics Competition, 1990)
[Solution 1] Since $2009=49 \cdot 41$, then $\sqrt{2009}=7 \sqrt{41}$, $\therefore \sqrt{x}+\sqrt{y}=\sqrt{41}+6 \sqrt{41}$ or $2 \sqrt{41}+5 \sqrt{41}$ or $3 \sqrt{41}+4 \sqrt{41}$,

i.e., $\sqrt{x}+\sqrt{y}=\sqrt{41}+\sqrt{1476}$ or $\sqrt{164}+\sqrt{1025}$ or $\sqrt{369}+\sqrt{656}$,
$\therefore(x, y)=(41,1476)$ or $(164,1025)$ or $(369,656)$.
Thus, $(x+y)_{\text {max }}=41+1476=1517$.
Therefore, the answer is $(B)$.
[Solution 2] Since $\sqrt{2009}=7 \sqrt{41}$, then $\sqrt{x}+\sqrt{y}=7 \sqrt{41}$.
$$
\therefore \quad x+y=(\sqrt{x}+\sqrt{y})^{2}-2 \sqrt{x y}=2009-2 \sqrt{x y} \text {, }
$$
Since $x, y$ are positive integers, then $(\sqrt{x y})_{\min }=\sqrt{41} \cdot 6 \sqrt{41}=246$.
$$
\therefore(x+y)_{\max }=2009-2 \cdot 246=1517 \text {. }
$$

Therefore, the answer is $(B)$.