For a given positive integer $n \ge 2$ let $a_n$ be the number of integers $k$ such that $\dfrac{k}{\pi(k)} = n$ where $\pi(x)$ is the prime counting function. The first few values of $(n,a_n)$ are
$$(2, 4), (3, 3), (4, 3), (5, 6), (6, 7), (7, 6), (8, 6), (9, 3), (10, 9)(11, 1), (12, 18),$$ $$(13, 11),(14, 12),(15, 21),(16, 3),(17, 10), (18, 33), (19, 31), (20, 32), (21, 24)$$
In the example above we see that for $n \ge 2, a_n \ge 1$. Intuitively this is expected because by the prime number theorem, $\dfrac{k}{\pi(k)} \sim \log k$. Hence as $k$ increases, the integer part of $\log k$ is expected to run through all positive integers after a certain point. However $a_k$ we notice that is not strictly increasing and $a_{11} = 1$. This brings the question:
Question: Is there a positive integer $n \ge 2$ for which $\dfrac{k}{\pi(k)} = n$ has no solution?
The answer is no, there is no such $n$. Observe that the fraction increases if and only if $\pi (k)$ stagnates. If $\frac{k}{\pi (k)} \leq n < \frac{k+1}{\pi (k+1)}$, then $\pi (k+1)=\pi (k)$ and thus $n\pi (k) -1<k\leq n\pi (k)$ with $k\in \mathbb{N}$, i.e. $k=n\pi (k)$.