$$
\begin{array}{l}
\text { Three. (25 points) Given the equation in } x \text { : } \\
x^{2}+\sqrt{a-2009} x+\frac{a-2061}{2}=0
\end{array}
$$

has two integer roots. Find all real values of $a$ that satisfy the condition.
Three, let the two integer roots of the equation be \(x_{1}\) and \(x_{2}\).
By Vieta's formulas, we have \(x_{1} + x_{2} = -\sqrt{a-2009}\) is an integer, and \(x_{1} x_{2} = \frac{a-2061}{2}\) is an integer.

Thus, the original equation is a quadratic equation with integer coefficients, and \(a-2009\) is a perfect square, with \(a\) being an odd number.

Assume \(a-2009 = k^2\) (where \(k\) is a natural number and \(k\) is even).
Then, \(a = k^2 + 2009\).
\[
\begin{array}{l}
\text{Also, } \Delta = a - 2009 - 4 \times \frac{a - 2061}{2} \\
= -(k^2 + 2009) - 2009 + 2 \times 2061 \\
= -k^2 + 104 \geqslant 0.
\end{array}
\]

Solving, we get \(-\sqrt{104} \leqslant k \leqslant \sqrt{104}\).
Since \(k \geqslant 0\), we have \(0 \leqslant k \leqslant \sqrt{104}\) (where \(k\) is even).
Thus, \(k = 0, 2, 4, 6, 8, 10\).
(1) When \(k = 0, 4, 6, 8\), \(\Delta\) is not a perfect square, and the original equation has no integer roots, which does not meet the problem's conditions;
(2) When \(k = 2\), \(\Delta = 100\), \(a = k^2 + 2009 = 2013\). In this case, the equation is \(x^2 + 2x - 24 = 0\), with roots \(x_1 = 4\), \(x_2 = -6\), which meets the problem's conditions;
(3) When \(k = 10\), \(\Delta = 4\), \(a = k^2 + 2009 = 2109\). In this case, the equation is \(x^2 + 10x + 24 = 0\), with roots \(x_1 = -4\), \(x_2 = -6\), which also meets the problem's conditions.

In summary, all real numbers \(a\) that satisfy the conditions are 2013 and 2109.
(Zhou Dongting, Wang Weilin, Experimental Middle School, Xishui County, Huanggang City, Hubei Province, 438200)