1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$
$-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$

where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}{2011!}\right|-2011 \\
>4024|q \cdot 2012!|-2011 \\
\geqslant 4024-2011=2013 .
\end{array}
$$

If $q=0$, then
$$
\begin{array}{l}
P(x)=x, \\
\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|=2011 .
\end{array}
$$

In summary, $\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|$'s minimum value is 2011 .