Well Calvin can't we have 2009,0 as the answer?I want my rating back......

Without loss of generality, let:

$\sqrt{a}=\sqrt{2009}-\sqrt{b}$

Squaring both sides will provide

$a=2009+b-2\sqrt{2009b}$

Notice that $2009=7^2 \times 41$

Rewrite ad

$a=2009+b-2\sqrt{7^2 \times 41 \times b}$

The $LHS$ must be positive integers and so do the $RHS$ . Thus, $\sqrt{7^2\times 41\times b}$ must be an integer. In other words, $7^2 \times 41 \times b$ is a perfect square.

Let $7^2 \times 41\times b =k^2$ for any integer $k$

$7^2$ is definitely a perfect square since the exponents is divisible by 2. $41$ is not a perfect square.

$b$ may have more than one solution, or even no solutions. Since $7^2 \times 41$ is not a perfect square, we have to multiply (the smallest solution) $41$ to make it a perfect square. But, there is a case that there exist a larger solution than $41$ . Thus, to make a guarantee of perfect square, the general solution of $b$ is

$b=41\times l^2$ for any $l$

Now, with a little logic, we find that

$2009>b>0$

Since $a$ will be any non-positive integers if $b>=0$

Now, rewrite as

$0<b<2009$

$0<41l^2<2009$

$0<l^2<49$

From inequality above, the solutions of $l$ is $1, 2, 3, 4, 5, 6$ and makes $b$ the solution $41 \times 1^2, 41 \times 2^2, 41 \times 3^2, 41 \times 4^2, 41 \times 5^2, 41 \times 6^2$

By trial and error, we have all possible $a$ solution. They are $1476, 1025, 656$

Finally, the sum of all possible values of $a$ is $1476+1025+656=3157$