The natural density of a set $S$ of natural numbers is defined to be $$ \lim_{n \to \infty} \frac{\#(S \cap \{1, 2, \cdots, n\})}{n}, $$ provided the limit exists.
Consider the set of natural numbers whose leading digit is $1$.
Its natural density does not exist (consider numbers of the form $2\cdot10^N$ and powers of $10$). Can this set be written as the intersection of two sets whose natural densities do exist?
No. Let $S$ be the set of all natural numbers with leading digit $1$. If $S\subseteq A\subseteq\mathbb N$, and if the (natural) density of $A$ exists, then the density of $A$ must be equal to $1$. It follows that $S$ can't be the intersection of a finite number of sets whose densities exist.
(Let $d$ be the density of $A$. To see that $d=1$, choose $n$ sufficiently large that the density of $A$ on the interval $[1,\ 10^n]$ and the density of $A$ on the interval $[1,\ 2\cdot10^n]$ are approximately equal to $d$; then $d\approx\frac{d+1}2$ so $d\approx1$.)
Likewise, a subset of $S$ whose density exists must have density zero, so $S$ can't be the union of a finite number of sets whose densities exist.