For those familiar with Latin squares terminology, I'll get straight to the point:
Q: For all $N \geq 2$, does there exists a critical set $C$ (for a Latin square of any finite order) such that deleting an entry from $C$ always results in $\geq N$ completions?
I'm guessing the answer is yes, but it doesn't strike me as obvious how to construct such critical sets. (Although, as always, I could be missing something.)
For those who aren't familiar (or who are rusty, or use different terminology)...
A partial Latin square is an $n \times n$ matrix with symbols from $\{1,2,\ldots,n\}$ and possibly some empty cells in which no symbol is repeated in a row or column.
A Latin square $L=(l_{ij})$ is said to be a completion of a partial Latin square $P=(p_{ij})$ if $l_{ij}=p_{ij}$ whenever cell $(i,j)$ in $P$ is nonempty.
A critical set is a partial Latin square $P$ that (a) admits a unique completion, and (b) admits $\geq 2$ completions after deleting any entry.
Here is an example of a critical set: $$\begin{array}{|cccc|} \hline 4 & 1 & \cdot & \cdot \\ \cdot & \cdot & \cdot & 2 \\ \cdot & 3 & \cdot & \cdot \\ \cdot & \cdot & 1 & \cdot \\ \hline \end{array}$$ which admits the unique completion $$\begin{array}{|cccc|} \hline 4 & 1 & 2 & 3 \\ 1 & 4 & 3 & 2 \\ 2 & 3 & 4 & 1 \\ 3 & 2 & 1 & 4 \\ \hline \end{array}.$$
If we delete the $1$ in the top row of the critical set, it admits the alternative completion $$\underbrace{\begin{array}{|cccc|} \hline 4 & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & 2 \\ \cdot & 3 & \cdot & \cdot \\ \cdot & \cdot & 1 & \cdot \\ \hline \end{array}}_{\text{exactly } 2 \text{ completions}} \xrightarrow{\text{also completes to}} \begin{array}{|cccc|} \hline 4 & \color{blue} 2 & 3 & 1 \\ 3 & 1 & 4 & 2 \\ 1 & 3 & 2 & 4 \\ 2 & 4 & 1 & 3 \\ \hline \end{array}.$$ It admits no other completions.
I would like, for arbitrary $N \geq 2$, to construct critical sets in which deleting any entry always results in $\geq N$ completions (if possible).