For $n \in \mathbb N_0$, let $L = {\downarrow}n$ as a sub-lattice of $(\mathbb N_0,\leq)$.
(Here, $a \leq b$ means $a$ divides $b$.)
For which elements $m \in L$ do we have a complement?
I know that $L$ is distributive and so if all elements are complemented, then $L$ is boolean.
I also know that this happens if and only if $n$ is square-free.
But it still may happen that $L$ is not boolean but some elements have complement (for example, $1$ and $n$ are complements of each other in $L$).
How to characterize those that are complemented?
$\DeclareMathOperator{\lcm}{lcm}$
If $n=0$, then $L = (\mathbb N_0, \leq)$ and the only complemented elements are $0$ and $1$ (notice that $1$ is the least element of this lattice and $0$ is the greatest).
Indeed, if $1 \neq m, m' \in \mathbb N$, then in $L$ $$m \vee m' = \lcm(m,m') \leq mm' < 0,$$ so the only element $x$ such that $m \vee x = 0$ is $x = 0$; but then $x \wedge 0 = x \neq 1$.
If $n \neq 0$, then $L$ is finite and $$n = p_1^{\nu_1} \cdots p_k^{\nu_k},$$ for some prime numbers $p_1, \dots, p_k$ and naturals $\nu_1, \dots, \nu_k$.
Then, for $m \in L$, $$m = p_1^{\mu_1} \cdots p_k^{\mu_k},$$ where the naturals $\mu_1, \dots, \mu_k$ are such that $\mu_i \leq \nu_i$.
If $\mu_i \in \{0,\nu_i\}$, then $m$ has the complement $m'$ defined by $$m' = p_1^{\delta_1} \cdots p_k^{\delta_k},$$ where $\delta_i = \nu_i - \mu_i$. Check it!
If for some $i$, $0 < \mu_i < \nu_i$, then if $$m \wedge m' = \gcd(m,m') = 1,$$ it must be $\delta_i = 0$; but in that case, you can easily check that $m \vee m' \neq n$, and so $m$ is not complemented.
Added. After a comment from OP, I'll add the following additional information.
As is it rightly pointed out in the question, if $n$ is square-free, then $L={\downarrow}n$ is a Boolean lattice.
This post generalizes that situation inasmuch as in that case the representation of $n$ as $$n=p_1^{\nu_1} \cdots p_k^{\nu_k}$$ is such that $\nu_i = 1$ for all $i$. It then follows that for $m \in L$ (that is, $m\mid n$), $$m = p_1^{\mu_1} \cdots p_k^{\mu_k}$$ where, given that $\mu_i \leq \nu_i = 1$, then either $\mu_i = 0$ or $\mu_i = 1 = \nu_i$, and so $m$ is complemented for all $m \in L$.