Mirimanoff Paradox The class $B$ is well-founded, if every descending chain of subclasses is finite. The class $B$ is not well-founded if there exists an infinite chain of classes $B_n $ such that 
$ \quad 
\dots \in B_n \in \dots \in B_2 \in B_1 \in B.
$

Paradox: The collection $\mathbf{WF}$ of well-founded sets (on a given set of atoms) is not itself a set. 

Let me try to elaborate on the formulation of the paradox. Let $\mathbf{WF}$ be the set of well-founded sets; then $\mathbf{WF}$ itself is well-founded, for were $\mathbf{WF} \ni x_0 \ni x_1 \ni x_2 \ni\dots\, $, then $x_0$ would be a not well-founded member of $\mathbf{WF}$, which is preposterous. Therefore  $\mathbf{WF} \in \mathbf{WF}$, so $\mathbf{WF}$ is not well-founded, i.e. contradiction.

Is this paradox only present in the naive set theory? If the answer is no, then how is it resolved?
I think this paradox is similar to Burali-Forti antinomy, but honestly I do not understand (the resolution of) the latter one as well. 
 A: Paradoxes of naive set theory are examples of contradictions apparent in the naive approach that "everything is a set", as formalized by the comprehension axiom[s].
Indeed if $\bf WF$ is a set then it is well-founded, then $\bf WF\in WF$, which is a contradiction to the well-foundedness. 
The resolution of this paradox, and indeed other paradoxes (like Burali-Forti that you mentioned), is that not every "property" defines a set. 
Some modern set theories (e.g. $\sf ZFC$) resolve this by requiring that properties establish subsets of previously constructed sets; other set theories (e.g. $\sf NF$) resolve this by requiring that the formulas themselves will have certain syntactic properties which prevent them from defining these classes.

And when all this has been said and done, I'd like to point out that the axiom of choice is necessary for the proof that well-foundedness is equivalent to the lack of infinite decreasing chains. The correct formulation should be that every non-empty set has an $\in$-minimal element (namely, if $x$ is not empty, then for some $y\in x$, $x\cap y=\varnothing$).
On the other hand, if you prove (without appealing to choice) that if $x$ is a well-founded set, then $\{x\}$ is a well-founded set, and that for every $x\in\bf WF$ satisfies $x\notin x$, from which we can conclude that $\bf WF\notin WF$.
