Let $y(x)$ be the unique solution of the differential equation
$$
\frac{\mathrm{d} y}{\mathrm{~d} x}=\log _{e} \frac{y}{x}, \quad \text { where } x>0 \text { and } y>0,
$$
with the initial condition $y(1)=2018$.
How many positive real numbers $x$ satisfy the equation $y(x)=2000$ ?
First, note that $y^{\prime}(x)>0$ whenever $y>x$ and $y^{\prime}(x)x, y(x)$ is decreasing for $y-1$. So $y(x)$ is a concave function.

Lemma 1. There exists $m>0$ such that $y(m)=m$.

Proof. In order to obtain a contradiction, assume otherwise and observe that the initial condition $y(1)=2018$ and the continuity of $y(x)$ imply that $y(x)>x$ for all $x>1$. Let
$$
k=\inf _{x \geq 1} \frac{y(x)}{x},
$$
and note we must have $k \geq 1$. Then there exist $0 \leq \epsilon0$ such that
$$
\frac{y(a)}{a}=k+\left.\epsilon \quad \Rightarrow \quad \frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=a}=\log _{e}(k+\epsilon) .
$$

The concavity of $y(x)$ then leads to the inequality
$$
y(x) \leq y(a)+(x-a) \log _{e}(k+\epsilon) \quad \Rightarrow \quad \frac{y(x)}{x} \leq \frac{y(a)}{x}+\frac{x-a}{x} \log _{e}(k+\epsilon) .
$$

By taking the limit as $x$ approaches infinity, this asserts the existence of a value of $x$ for which
$$
\frac{y(x)}{x} \leq \log _{e}(k+\epsilon)+\frac{1}{2}-1$ and the fact that $\epsilon0$ such that $y(m)=m$.

For the value of $m$ for which $y(m)=m$, we have $y(x)$ is increasing for $xm$, so $y(m)$ is a unique global maximum. Combined with the concavity of $y(x)$, this implies that $y(x)$ is only defined on the open interval $(0, c)$ for some $c>0$. Hence, there is exactly one solution to $y(x)=2000$ on the interval $(m, c)$.
Now let us show that there is no solution to $y(x)=2000$ on the interval $(0, m)$. Since $y(x)$ is increasing on this interval, it suffices to show that $\lim _{x \rightarrow 0^{+}} y(x)>2000$. For $02018-\int_{0}^{1}\left(\log _{e} 2018-\log _{e} x\right) \mathrm{d} x=2017-\log _{e} 2018>2000 .
$$
(This final inequality can be obtained from observing that $\log _{e} 2018<\log _{2} 2048=$ 11.) Therefore, there is exactly one positive real number $x$ that satisfies the equation $y(x)=2000$.