Reverse Skolem's paradox By using the Löwenheim–Skolem theorem & Mostowski collapse, in every model $V$ of $ZF+Con(ZF)$ there is a countable transitive set $M$ such that $(M,\in_M) \models ZF$. Is the following "converse" true?

In every model $V$ of $ZF$ and every transitive set $M \in V$ such that $(M,\in_M) \models ZF$, there exists a transitive set $N \in V$ such that:
  
  
*
  
*$M \in N$
  
*$(N,\in_N) \models ZF$
  
*$M$ is countable inside $N$

 A: The answer to your question is no. For example, suppose $\kappa$ is inaccessible and $M = V_{\kappa}$. Then $M$ is a model of $\sf ZF$ but $M$ cannot be countable in $N$ as it is not countable in $V$.
Of course there will always be a set generic extension of $V$ containing such an $N$ (just generically add a bijection from $M$ to omega)
A: Let me point out that a close variant of the question has a
positive answer. Namely,
Theorem. Every model of set theory $M\models\text{ZFC}$ has an
elementary extension $M\prec M^*$ such that there is a model
$N\models\text{ZFC}$ inside of which $\langle
M^*,\in^{M^*}\rangle$ is a countable transitive model.
Proof. Fix any model $\langle
M,\in^M\rangle\models\text{ZFC}$. First, I claim that there is a
model $W\models\text{ZFC}$ with a cardinal $\delta$ such that
$$M\prec V_\delta^W\prec W.$$ To see this, consider the theory
$T$, in the language of set theory augmented with a constant
symbol $\delta$, which asserts the elementary diagram of $M$,
together with the assertions that $\delta$ is an ordinal and every
element of $M$ is in $V_\delta$ and also the scheme $V_\delta\prec
V$, asserting that $\delta$ is a correct cardinal. The
reflection theorem shows that every finite subset of this theory
is consistent, and so $T$ has a model $W$, which means that
$M\prec V_\delta^W\prec W$, as desired.
Second, we will go to a further extension in which (the image of)
$V_\delta^W$ becomes countable. Specifically, inside $W$ consider
the forcing that makes $\delta$ countable. Let $U\in W$ be an
ultrafilter on the corresponding complete Boolean algebra
$\mathbb{B}$, and let $j:W\to \check W_U\subset W^{\mathbb{B}}/U$
be the Boolean ultrapower embedding. This embedding and these
models exist inside $W$, without any actual forcing; I am just
taking a quotient by an ultrafilter in $V$. You can find further
explanation in my paper, Well-founded Boolean ultrapowers as
large cardinal embeddings.
Note that $j:V_\delta^W\to j(V_\delta^W)=[{\check V}_\delta^W]_U$ is an elementary
embedding, and furthermore, $j(V_\delta^W)$ is a countable
transitive set inside $W^{\mathbb{B}}/U$. So let
$M^*=j(V_\delta^W)$ and $N=W^{\mathbb{B}}/U$, and we have achieved
the statement of the theorem. QED
The theorem applies to the case $M=V_\kappa$ in Nate's answer as
follows: the model $M^*$ will be $\omega$-nonstandard (as in
Asaf's answer), with $\kappa$ many natural numbers, and so the
former contradiction of Nate's answer does not engage. The
uncountable model $M$ is thought countable inside $N$, since $N$
has a lot of natural numbers.
One can actually improve the theorem to have
$N\models\text{ZFC}+V=L$, since every countable model is a
transitive model inside a model of $V=L$. This and many other
similar results are proved and discussed in my paper, A
multiverse perspective on the axiom of constructibility.
A: While the original question has been answered, remember that Skolem's paradox says that if $M$ is a countable model of set theory, then it knows only about countably many real numbers.
The reverse Skolem paradox, if so, is the following situation that can occur:

There $(M,E)\models\sf ZF$ such that $\{x \mid M\models x\text{ is an integer}\}$ is uncountable. (Assuming that there is a model of $\sf ZF$ to begin with, but let's dispense of that issue for now.)

The point is that $M$ thinks that an uncountable set is in fact countable. This $M$ cannot be transitive (or well-founded), and obtaining it is a simple exercise in compactness or ultrapowers. This shows that not only that "uncountable" is relative and internal to the model, but also "countable", and in fact in this model $M$, even "finite" is relative. 
Since $M$ will think about certain infinite set that they are finite (e.g. various bounded sets of "natural numbers" in $M$ will contain infinitely many objects, but $M$ will still think they are finite because $M$ has a twisted view as for what is finite and countable).
