When do infinitary compactness numbers exist? For a logic $\mathcal{L}$, let the compactness number of $\mathcal{L}$ (if it exists) be the least $\kappa$ such that every $(<\kappa)$-satisfiable $\mathcal{L}$-theory is satisfiable. Note that there is no restriction here on the cardinality of the language of the theory in question.
For example, an uncountable cardinal $\kappa$ is strongly compact iff it is the compactness number of its own infinitary logic $\mathcal{L}_{\kappa,\kappa}$; more interestingly, Magidor showed that $\mathsf{SOL}$ has a compactness number iff there is an extendible cardinal, in which case its compactness number is the least extendible cardinal.
My question is:

What is the strength of "For every $\kappa$, the compactness number of $\mathcal{L}_{\kappa,\kappa}$ exists?"

EDIT: Originally I said that I didn't know anything relevant, but I just noticed that one of the suggested related questions is very relevant, namely this one: there it is shown for example that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ already implies the existence of a measurable cardinal, or more technically that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ is equivalent to the existence of an $\omega_1$-strongly compact cardinal. A natural guess based on that is that the principle in question is equivalent to "For every $\kappa$ there is a $\kappa$-strongly compact cardinal," but I haven't had a chance to read through the argument in detail so I'm not too confident here.
 A: The compactness number for $\mathcal L_{\kappa,\kappa}$ is equal to the least $(\kappa,\infty)$-strongly compact cardinal. A cardinal is $(\kappa,\infty)$-strongly compact if for every set $X$, there is a $j : V\to M$ such that $\text{crit}(j)\geq \kappa$, and $j[X]$ can be covered by and element of $M$ of $M$-cardinality less than $j(\delta)$. I sketch a proof at the end because I don't know the reference.
But first: it follows easily that your hypothesis is equivalent to the existence of a proper class of almost strongly compact cardinals, which are (resp. should be) defined to be cardinals $\kappa$ such that for all $\gamma < \kappa$ every $\kappa$-complete filter can be extended to a $\gamma$-complete (resp. $\gamma^+$-complete) ultrafilter. Whether this is equivalent to the existence of a proper class of strongly compact cardinals is an open question.
The true consistency strength is probably a proper class of supercompacts: all three of these hypotheses are equivalent under the Ultrapower Axiom. There is some evidence that the equivalence between a proper class of almost strong compacts and a proper class of strong compacts is a theorem of ZFC: the first almost strongly compact cardinal above an ordinal $\gamma$ is either strongly compact or else has countable cofinality (although the truth is I needed a little SCH to handle the case $\gamma = 0$). This is in Some combinatorial properties of Ultimate $L$ and $V$.
Now the proof. In one direction, you show that $\mathcal L_{\kappa,\kappa}$ is $\delta$-compact for any $\kappa$-strongly compact $\delta$. Fix a $\delta$-consistent theory $T$ in the signature $\tau$. Cover $j[T]$ by a theory $S\subseteq j(T)$ in $M$ of $M$-cardinality less than $j(\delta)$. You get a model $\mathfrak A$ of $S$ in $M$ by $j(\delta)$-consistency of $j(T)$. Take the reduct of $\mathfrak A$ to $j[\tau]$. This is essentially a model of $T$: more precisely, $j : T \to j[T]$ is an isomorphism of $\mathcal L_{\kappa,\kappa}$-theories because $\text{crit}(j)\geq \kappa$.
Conversely, if $\delta$ is the compactness number of $\mathcal L_{\kappa,\kappa}$, then for any set $X$ and any $\delta$-complete filter base $\mathcal B$ on $X$, you can build a $\delta$-consistent theory whose models are $\kappa$-complete ultrafilters on $X$ extending $\mathcal B$. (A $\delta$-complete filter base is a family of sets such that the intersection of any ${<}\delta$-sized subfamily is nonempty.)
The signature has constants for all subsets of $X$ along with a predicate $W$. The theory contains the axiom "$W(A)$" for each $A\in \mathcal B$ and the axiom "If $W(\bigcup \mathcal P)$, then $\bigvee_{A\in \mathcal P}W(A)$" for every partition $\mathcal P$ of $X$ with $|\mathcal P| < \kappa$. The theory is $\delta$-consistent since if one takes a set $\mathcal A\subseteq P(X)$ of cardinality less than $\delta$, one obtains a model of the axioms in the signature restricted to constants from $\mathcal A$ by letting $W$ be the principal ultrafilter concentrated at $x\in \bigcap(\mathcal A\cap \mathcal B)$.
It follows that for any set $X$, there is a $\kappa$-complete ultrafilter on $P_{\delta}(X)$ extending the filter base $\langle A_x \rangle_{x\in X}$ where $A_x = \{\sigma \in P_\delta(X): x\in \sigma\}$. Such an ultrafilter is, by definition, fine. The associated ultrapower embedding $j : V\to M$ has critical point at least $\kappa$ and closure under $\kappa$-sequences by $\kappa$-completeness. Finally $\text{id}_\mathcal U$ is a cover of $j[X]$ by fineness, and $\text{id}_\mathcal U$ has $M$-cardinality less than $j(\delta)$ since it is an element of $j(P_{\delta}(X))$ by the definition of $M$-membership. So $\delta'$ is $(\kappa,\infty)$-strongly compact. But it is not too hard to show that the least $(\kappa,\infty)$-strongly compact cardinal is a limit cardinal, so $\delta$ must be $(\kappa,\infty)$-strongly compact.
