I found a mention in F. A. Muller's 2004 paper that adding 84 inaccessibles to ZFC makes the theory inconsistent if it is consistent with 17 ones, but adding instead 49 can be done consistently, although the resulting theory has no well-founded models. I find this intriguing. Is this true? If so, can someone give an intuitive sketch of a proof?
Muller, F. A., Deflating Skolem, Synthese 143, No. 3, 223-253 (2005). ZBL1083.03015.
This is an utterly bizarre claim, and quite frankly gibberish as written. That said, there is an interesting nontrivial result which has vaguely the same "look," and may have been what the author intended (certainly it fits well into the claim being made in the article).
First, while the paper is freely available on the author's website, it's worth noting that the OP has gotten the claim completely correctly; a direct quote is
This seems thoroughly garbled to me, and I suspect the author was just being careless here. The point they're making in this section is that there is a plethora of odd principles easily deduced from general metatheorems, and this is definitely true. But I can't parse what they've written.
The following is a precise and true statement; I suspect the author had something like this in mind. Let $T_k$ be the theory ZFC + "There are (at least) $k$ inaccessible cardinals." Then:
(And more generally for $k<m<n$ the analogous claim $(*)_{k,m,n}$ is true.)
The proof of a claim like this is a combination of basic set-theoretic arguments and Godel's incompleteness theorem.
As a first step, note that by Godel's theorem if $T_k$ is consistent then so is $T_k+\neg Con(T_k)$. Meanwhile, for $k<n$ we have $T_n\vdash Con(T_k)$. Taking $k=49, n=84$ we get that if $T_{49}$ is consistent then so a fortiori is $T_{49}+\neg Con(T_{84})$.
This next part is a bit technical:
Of course, claims of the form $(*)_{a,b,c}$ above are more subtle. There we have three levels of consistency (transitive model, some model, no model), and this makes this more complicated; we need to do a bit more work.
If there is a model of $T_{49}+Con(T_{49})$, then there is a model of $$T_{49}+Con(T_{49})+\neg Con(T_{84}).$$ Let $M$ be such a model, and let $\alpha$ be the least ordinal (in the sense of $M$) such that $(L_\alpha)^M\models T_{18}$.
Let $N=(L_\alpha)^M$. $N$ and $M$ have the same natural numbers, so they agree about consistency statements. In particular, $N$ thinks that $T_{49}$ is consistent and $T_{84}$ is inconsistent. Meanwhile, $N$ also has a transitive model of $T_{17}$. But we because we "cut off" $M$ at the first point where we could get $18$ inaccessibles (this also uses the "minimality" of $L$), $N$ does not think that there is a transitive model of $T_{49}$.
So from a model of $T_{49}+Con(T_{49})$ we can build via incompleteness a model of $T_{49}+Con(T_{49})+\neg Con(T_{84})$, and in such a model an appropriate initial segment of the $L$ hierarchy will be in turn satisfy $T_{49}+Con(T_{49})+\neg Con(T_{84})+$ "$T_{49}$ has no transitive model" (note that $T_{49}$ already proves that $T_{17}$ has a transitive model, so we don't need to add that separately).
As an aside, let me say a bit about the ordinal $\alpha$ above. I claim that $\alpha$ is countable. The countability of $\alpha$ is in fact a special case of a stronger fact:
This is a consequence of the Lowenheim-Skolem and Condensation theorems:
Suppose $L_\alpha\models T$. By the Lowenheim-Skolem theorem there is some countable $A\preccurlyeq L_\alpha$; this $A$ also satisfies $T$ (by elementarity).
Let $B$ be the Mostowski collapse of $A$. Since $B$ is isomorphic to $A$ (more precisely: $(A,\in\upharpoonright A)\cong (B,\in\upharpoonright B))$ we have $B\models T$.
But by Condensation, $B=L_\gamma$ for some $\gamma$. Since $B$ is countable, $\gamma$ is also countable, and we're done.
Note that this looks a lot like Skolem's paradox (that if ZFC has a model, it has a countable model). Indeed, this is really just a stronger version of Skolem's paradox in the more specific setting of ZFC + V=L.