3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ .
3.2 .

Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b$ and $a-b$ have the same parity, the minimum value of $a-b$ is 2.
Therefore, the minimum value of $y$ is 2.