Problem 3. Determine the real number $x>0$ and the non-zero natural number $n$ for which

$$
[x]+\left\{\frac{1}{x}\right\}=1.005 \cdot n
$$
Solution. The equation is written as $[x]+\left\{\frac{1}{x}\right\}=\frac{201}{200} n=n+\frac{n}{200}$. Denoting by $q$, respectively $r$, the quotient and the remainder of the division of $n$ by 200, it follows that

$$
[x]+\left\{\frac{1}{x}\right\}=n+q+\frac{r}{200}
$$

The integer part of the expression on the left side is equal to $[x]$, and the integer part of the expression on the right side is $n+q$, so $[x]=201 q+r$ and $\left\{\frac{1}{x}\right\}=\frac{r}{200} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots 2$ points

If $x>1$, then $0<\frac{1}{x}<1$, so $\left\{\frac{1}{x}\right\}=\frac{1}{x}$. It follows that $\frac{1}{x}=\frac{r}{200}$, so $r \neq 0$ and $x=\frac{200}{r}$.

Using the fact that $[x] \leq x<[x]+1$, it follows that $201 q+r \leq \frac{200}{r}<201 q+r+1$, hence $201 q r+r^{2} \leq$ $200<201 q r+r(r+1)$. Since $q, r \in \mathbb{N}$, the inequalities above cannot hold for $q r \geq 1$, so $q r=0$. Since $r \neq 0$, it follows that $q=0$, so $r^{2} \leq 200<r(r+1)$, from which we get $r=14, n=14$ and $x=\frac{100}{7}$