Definition. $x \in \mathbb{R}$ is a condensation point of a subset $A \subseteq \mathbb{R} \iff$ the intersection of every neighbourhood of $x$ with $A$ is uncountable.
Can one construct a subset $N \subseteq \mathbb{R}$ that has lebesgue measure zero, such that every point in $\mathbb{R}$ is a condensation point of $N$ ?
Let $I_n = [a_n, b_n]$ be an enumeration of the set of intervals with rational endpoints. For each $n$ let $C_n \subseteq I_n$ be a cantor set contracted to $I_n$. To do so, if $C$ is the cantor set and $l_n$ is the length of $I_n$, we let $C_n = (l_nC) + a_n$ where $l_nC = \{l_n x\;|\;x \in C\}$. Then, recall that the cantor set has Lebesgue measure 0, denoted $|C| = 0$.
Then, by countable subadditivity we have that $$\left| \bigcup_{n \in \mathbb{N}} C_n \right| \leq \sum_{n \in N} |C_n| = \sum_{n \in N} |C| = 0.$$ Moreover, if $p \in \mathbb{R}$ and $(p - \epsilon, p + \epsilon)$ is contained in some neighborhood of $p$, we know that there are distinct rational numbers $q, q' \in (p - \epsilon, p + \epsilon)$ with $q < q'$. Then, $[q, q'] = I_k$ for some $k$ and $[q, q'] = I_k \subseteq (p - \epsilon, p + \epsilon)$. By construction, $C_k \subseteq I_k \subseteq (p - \epsilon, p + \epsilon)$. Since $C_k$ is in bijection with $C$, which is uncountable, we find that $p$ is a condensation point of $\bigcup_{n \in \mathbb{N}} C_n $.
Hence the set $\bigcup_{n \in \mathbb{N}} C_n $ is what we're looking for.