I am trying to find what the list chromatic number of $K_{n,n}$ is for my homework but I have not been able to do so.
The problem is somewhat analogous to the following: Given $l\in \mathbb Z^+$ what is the first $n$ such that $K_{n,n}$ is not always colorable with $l$-lists.
In my homework we were just asked to prove that such an $n$ exists, this is easy because for $n = \binom{2l}{l}$ we can consider taking the ground set $\{1,\dots,2l\}$ and assign each $l$-subset of it to one vertex on each side. It follows that if this list assignation gave a coloring there would be at least $l+1$ colors used on each side, which is a contradiction as there would be a color used on both sides.
Now, we can see that $K_{n,n}$ is not list colorable if and only if there exists a ground set size $b$ such that there is a covering family consisting of $n$ subsets of $[b]$ of size $b-l$ that covers each $\lfloor b/2 \rfloor$-subset. I am wondering, given $l$ what the best $b$ is but I haven't been very succesful, perhaps I am missing something, I honestly haven't even been able to see if the best $b$ is always $2l$ ( which gives $n=\binom{2l}{l}$).
Thank you kindly and best regards.
By using the information provided in the comments, this problem is open.
The problem of finding for each $l$ the first graph $K_{g(l),g(l)}$ that is no $l$-list colorable is closely related to finding the first $f(l)$ such that there is a bipartite graph of order $f(l)$ that is not $l$-list colorable.
Clearly one has $g(l) \leq f(l) \leq 2g(l)$.
It follows good bounds on $g(l)$ lead to good bounds on $f(l)$.
The bound mentioned is $2^{l-1} < f(l) < l^2 2^{l+2}$