3. Given that $x, y$ are integers, and
$$
y=\frac{4012}{\sqrt{x+2005}-\sqrt{x-2007}} \text {. }
$$

Then the maximum value of $y$ is $\qquad$ .
3.2006 .

Rationalizing the denominator of the given equation, we get
$$
y=\sqrt{x+2005}+\sqrt{x-2007} \text {. }
$$

Let $m=\sqrt{x+2005}, n=\sqrt{x-2007}$.
Then $y=m+n, m^{2}=x+2005, n^{2}=x-2007$ $(m>n \geqslant 0)$, obviously $m, n$ are integers.
Eliminating $x$ yields $m^{2}-n^{2}=4012$, that is,
$$
(m+n)(m-n)=2^{2} \times 17 \times 59 \text {. }
$$

Since $m+n$ and $m-n$ have the same parity, $m+n$ and $m-n$ are both even.

Therefore, the minimum value of $m-n$ is $2$, and the maximum value of $m+n$ is $2 \times$ $17 \times 59=2006$.