Example 2 The equation $2 x^{2}+5 x y+2 y^{2}=2007$ has $\qquad$ different integer solutions.
(2007, National Junior High School Mathematics League Sichuan Preliminary Competition)
The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2007 \text {. }
$$

Since $x$ and $y$ are integers, without loss of generality, assume $x \leqslant y$, so,
$$
2 x+y \leqslant x+2 y \text {. }
$$

Notice that, $31[2 x+y)+(x+2 y)]$.
Thus, from equation (1), we get the system of equations
$$
\left\{\begin{array} { l } 
{ 2 x + y = 3 , } \\
{ x + 2 y = 6 6 9 ; }
\end{array} \left\{\begin{array}{l}
2 x+y=-669, \\
x+2 y=-3 .
\end{array}\right.\right.
$$

The integer solutions are

When $x \geqslant y$, the original equation has two more sets of integer solutions
$$
\left\{\begin{array} { l } 
{ x = 4 4 5 , } \\
{ y = - 2 2 1 , }
\end{array} \left\{\begin{array}{l}
x=221, \\
y=-445 .
\end{array}\right.\right.
$$

In summary, the original equation has 4 sets of integer solutions.