1. Find the number of natural solutions to the equation

$$
\left[\frac{x}{2010}\right]=\left[\frac{x}{2011}\right]+1
$$
5.I.1. Let's represent the number $x$ in the form $x=2011 k+d$, where $d=0,1, \ldots 2010$. Since $\left[\frac{x}{2010}\right]=\left[\frac{x}{2011}\right]+1=k+1$, then $k+1 \leqslant \frac{x}{2010}<2011$, from which

$$
2010 k+2010 \leqslant 2011 k+d<2010 k+4020
$$

thus, $2010 \leqslant k+d<4020$, or $2010 \leqslant k+d \leqslant 4019$. Therefore, for each of the 2011 values of $d$ there are 2010 values of the number $k$ such that the number $x=2011 k+d$ is a solution to the given equation. Hence, it has $2010 \cdot 2011$ solutions.

Answer: $2010 \cdot 2011$.