4. Equation
$$
\begin{array}{l}
\frac{x_{1} x_{2} \cdots x_{2010}}{x_{2011}}+\frac{x_{1} x_{2} \cdots x_{2009} x_{2011}}{x_{2010}}+\cdots+ \\
\frac{x_{2} x_{3} \cdots x_{2011}}{x_{1}}=2011
\end{array}
$$

The number of different ordered integer solutions $\left(x_{1}, x_{2}, \cdots, x_{2011}\right)$ is
4. $2^{2010}$.

The equation can be transformed into
$$
x_{1} x_{2} \cdots x_{2011}\left(\frac{1}{x_{2011}^{2}}+\frac{1}{x_{2010}^{2}}+\cdots+\frac{1}{x_{1}^{2}}\right)=2011 \text {. }
$$

From this, we know that $x_{1} x_{2} \cdots x_{2011}>0$.
$$
\begin{array}{l}
\text { Also, } \frac{2011}{x_{1} x_{2} \cdots x_{2011}}=\frac{1}{x_{2011}^{2}}+\frac{1}{x_{2010}^{2}}+\cdots+\frac{1}{x_{1}^{2}} \\
\geqslant 2011 \sqrt{\frac{1}{\left(x_{1} x_{2} \cdots x_{2011}\right)^{2}}},
\end{array}
$$

This can be transformed to $\left(x_{1} x_{2} \cdots x_{2011}\right)^{2009} \leqslant 1$, so,
$\left|x_{1} x_{2} \cdots x_{2011}\right| \leqslant 1$.
Since $x_{1}, x_{2}, \cdots, x_{2011}$ are integers, and $x_{1} x_{2} \cdots x_{2011} >0$, we get $x_{1} x_{2} \cdots x_{2011}=1$.
Thus, $x_{i}= \pm 1$, and there must be an even number of $x_{i}$ that are -1.
When $x_{1}, x_{2}, \cdots, x_{2011}$ have $2 k(k=0,1, \cdots, 1005)$ of -1, the number of solutions to the equation is $\mathrm{C}_{2011}^{2 k}$.
Therefore, the number of solutions to the equation is
$$
\mathrm{C}_{2011}^{0}+\mathrm{C}_{2011}^{2}+\mathrm{C}_{2011}^{4}+\cdots+\mathrm{C}_{2011}^{2010}=2^{2010}
$$