Task B-1.1. How many negative integers $x$ satisfy $\frac{x-2011}{x+2012} \leq 1$?
Solution.

$$
\begin{aligned}
& \frac{x-2011}{x+2012} \leq 1 \\
& \frac{x-2011}{x+2012}-1 \leq 0 \\
& \frac{x-2011-x-2012}{x+2012} \leq 0 \\
& \frac{-4023}{x+2012} \leq 0 \\
& x+2012>0 \\
& x>-2012
\end{aligned}
$$

There are 2011 negative integers that satisfy the given inequality.