We consider a relation $|$ on set $\mathbb N\setminus\{0\}$
Find set of minimal elements in set $\langle \mathbb N\setminus\{0\},|\rangle$
Prove that in set $\langle \mathbb N\setminus\{0\},|\rangle$ there is no maximal element.
A minimal element of a subset S of some preordered set is defined dually as an element of S that is not greater than any other element in S. Same situation with maximal element.
Yes, | is divisibility.
For me, set of minimal elements is just {1} because every number greater than 1 is divided by 1 and to prove that there are no maximal elements in this set, then I'd take n∈N and show that n is not maximal element, it means that there exists element greater in |(divisibility) relation. For example 2n. Or better proof is by assumption that there are maximal elements and came to contradiction?
(Converted from a previous comment, which the OP says was helpful.)
The argument that no $n$ can be a maximal element (because $n\mid 2n$ and $n\ne 2n$) is good.
The argument "$1$ is the minimal element because it divides all numbers" is a bit incomplete: what you said implies that no other number is minimal, but doesn't fully prove that $1$ is minimal. You should've added that the only divisor of $1$ is $1$ itself (which is the true meaning of "$1$ is minimal" here).
To see the difference better, consider the example $\langle\mathbb N\setminus\{0,1\},\mid\rangle$, where the set of minimal elements is... the set of all prime numbers. (Why?)