checking Zorn's lemma on an example
Your misunderstanding is visible right in the question (emphasis by me):
What is the maximum element of A with respect to the partial order relation?
Zorn's lemma does not guarantee you a maximum, but only a maximal element. Those two are not equivalent: There can be only one maximum, but there can be many maximal elements.
Here are the relevant definitions:
A maximum is an element $m$ such that for all other elements $n$ you have $n\le m$.
A maximal element is an element $m$ such that there is no element $n$ with $m<n$.
While for total orders they are equivalent, for partial orders they are not. In particular, a set can have at most one maximum, but many maximal elements.
Now obviously a maximum is a maximal element, and if a maximum exists, it is the only maximal element. But a maximal element need not be a maximum, and in particular, if a set has more than one maximal element, it cannot have a maximum.
Now to your set $A=\{1,\ldots,100\}$ partially sorted by divisibility:
Theorem: An element $n\in A$ is a maximal element in the divisibility order if and only if $n>50$ in the standard order.
Proof:
Assume $n>50$. Then the smallest (in the standard order) number of which it is a proper divisor is $2n$, but $2n>100$ is not an element of $A$, nor is any number that is larger (in the standard order). Therefore in the divisibility order no element of $A$ is larger than $n$, that is, $n$ is a maximal element in the divisibility order.
Conversely, assume $n\le 50$. Then $2n\le 100$, therefore $2n\in A$, but $n|2n$ and therefore $n$ is not a maximal element in the divisibility order.
Edit:
I just noticed that you also included $0$ in your set $A$, which I previously missed.
In that case, $A$ has indeed a maximum in the divisibility order, and that maximum is $0$:
By the definition you gave, $n|0$ iff there exists a $k\in\mathbb N$ such that $0=kn$. Obviously there exists such a $k$ for any $n$, namely $k=0$. Therefore any number divides $0$, and thus $0$ is the maximum (and the unique maximal element) of $A$.
Let's denote your relation by $\preceq,$ i.e. $a \preceq b \iff a | b. $ A maximal element of $A$ is $100$ (assuming $0 \notin A$, otherwise $0$ is maximal). Assume $a \in A$ with $100 \preceq a,$ then $100 | a.$ But that means that $a=100,$ so there is no bigger element in $A$ with respect to that relation.