Three. (25 points) Does there exist a right triangle with a hypotenuse of $\sqrt{2009}$ and two legs of integer lengths? If it exists, find the lengths of the two legs; if not, explain why.

---

Translated as requested, maintaining the original format and line breaks.
Three, assuming there exists such a right-angled triangle. Let the lengths of the two legs be \( x \) and \( y \). Then \( x^2 + y^2 = 2009 \).

Since 2009 is an odd number, \( x \) and \( y \) must be one odd and one even. Without loss of generality, let \( x = 2k \) and \( y = 2l + 1 \). Then
\[ 4k^2 + 4l(l+1) = 2008 \Rightarrow k^2 + l(l+1) = 502 \].
Since \( l(l+1) \) is even, \( k \) must be even. Let \( k = 2n \) and \( l(l+1) = 2g \). Then
\[ 4n^2 + 2g = 502 \Rightarrow 2n^2 + g = 251 \]
\[ \Rightarrow 2n^2 = 251 - g > 0 \Rightarrow 0 < n \leq 11 \]
\[ \Rightarrow n = 1, 2, \cdots, 11 \].

By verification, \( n = 1, g = 249, 2g = 498 \);
\[ \begin{array}{l}
n = 2, 2g = 486; n = 3, 2g = 466; \\
n = 4, 2g = 438; n = 5, 2g = 402; \\
n = 6, 2g = 358; n = 7, 2g = 306; \\
n = 8, 2g = 246; n = 9, 2g = 178; \\
n = 10, 2g = 102; n = 11, 2g = 18.
\end{array} \]

Since the last digit of \( l(l+1) \) is 2, 0, or 6, \( l(l+1) \) could be 486, 466, 402, 306, 246, or 102.

By calculation, only when \( l(l+1) = 306 \), \( l \) is an integer, at this time, \( l = 17 \).

Therefore, when \( x = 2k = 4n = 28 \) and \( y = 2 \times 17 + 1 = 35 \), the right-angled triangle satisfying the condition exists. At this time, the lengths of the two legs of the triangle are 28 and 35.