4. If $x, y \in[-2006,2006]$, and satisfy
$$
\begin{array}{l}
1+\cos ^{2}(2 x+3 y-1) \\
=\frac{x^{2}+y^{2}+2(1+x)(1-y)}{x-y+1},
\end{array}
$$

then the minimum value of $xy$ is
4. $\frac{1}{25}$.

Since $1+\cos ^{2}(2 x+3 y-1)>0$, we have
$$
\begin{array}{l}
\frac{x^{2}+y^{2}+2(1+x)(1-y)}{x-y+1} \\
=\frac{(x-y+1)^{2}+1}{x-y+1} \\
=x-y+1+\frac{1}{x-y+1}>0 .
\end{array}
$$

Thus, $x-y+1+\frac{1}{x-y+1} \geqslant 2$.
Also, $1 \leqslant 1+\cos ^{2}(2 x+3 y-1) \leqslant 2$, so, $1+\cos ^{2}(2 x+3 y-1)=2$, and $x-y+1=1$.
Therefore, $2 x+3 y-1=k \pi$, and $x=y$.
Hence, $x=y=\frac{k \pi+1}{5}(k \in \mathbf{Z})$.
Thus, $x y=\left(\frac{k \pi+1}{5}\right)^{2} \geqslant\left(\frac{1}{5}\right)^{2}=\frac{1}{25}$.
The equality holds if and only if $x=y=\frac{1}{5}$.