Let $D(x) = \{y \in \mathbb N : y\text{ is a divisor of } x\}$ and let the relation $\Sigma$ be defined as follows:
$\begin{aligned}x \Sigma y \Leftrightarrow D(x) \subseteq D(y) \end{aligned}$
check $\Sigma$ is a partial order, if it is total and determine $\min(\mathbb N, \Sigma)$ and $\max(\mathbb N, \Sigma)$.
In order to prove $\Sigma$ to be a partial order I need to gain proof of its reflexivity, anti-simmetry and transitivity, so:
Reflexivity
$x \Sigma x \Leftrightarrow D(x) \subseteq D(x)$
$\{x\in \mathbb N : x \text{ is a divisor of } x\} \subseteq \{x\in \mathbb N : x \text{ is a divisor of } x\}$
Anti-symmetry
$x \Sigma y \land y\Sigma x \Rightarrow x= y$
$\{y \in \mathbb N : y \text{ is a divisor of } x\} \land \{x \in \mathbb N : x \text{ is a divisor of } y\}$
for the fundamental theorem of arithmetic $x = y$
Transitivity
$x \Sigma y \land y\Sigma z \Rightarrow x \Sigma z$
$\{y\in \mathbb N : y \text{ is a divisor of } x\} \land \{z \in \mathbb N : z \text{ is a divisor of } y\} \Rightarrow \{z \in \mathbb N : z \text{ is a divisor of } x\}$.
I should prove if $\Sigma$ is a total order, but I have no idea how I could do that, can it involve the fact that element $0 \in \mathbb N$ doesn't belong to this relation because $x\Sigma 0$ would mean $\{0 \text{ is a divisor of } x\}$ which is not possible? And my gut feeling tells me that $\max(\mathbb N, \Sigma) = 1$ and $\nexists\min(\mathbb N, \Sigma)$ but I don't know if I am right. Will you please give me a nudge in the right direction?