Problem: Let $n$ be a nonnegative integer.
Determine the number of ways to choose sets
$S_{ij} \subseteq \{1, 2, \dots, 2n\}$,
for all $0 \le i \le n$ and $0 \le j \le n$
(not necessarily distinct), such that
\begin{itemize}
  \ii $|S_{ij}| = i+j$, and
  \ii $S_{ij} \subseteq S_{kl}$ if $0 \le i \le k \le n$
  and $0 \le j \le l \le n$.
\end{itemize}

Solution: The answer is $(2n)! \cdot 2^{n^2}$.
First, we note that
$\varnothing = S_{00} \subsetneq S_{01} \subsetneq \dots \subsetneq S_{nn}
= \left\{ 1, \dots, 2n \right\}$
and thus multiplying by $(2n)!$
we may as well assume $S_{0i} = \left\{ 1, \dots, i \right\}$
and $S_{in} = \left\{ 1, \dots, n+i \right\}$.
We illustrate this situation by placing the sets in a grid,
as below for $n = 4$;
our goal is to fill in the rest of the grid.
\[
  \begin{bmatrix}
    1234 & 12345 & 123456 & 1234567 & 12345678\\
    123 \\
    12 \\
    1 \\
    \varnothing
  \end{bmatrix}
\]
We claim the number of ways to do so is $2^{n^2}$.
In fact, more strongly even the partial fillings
are given exactly by powers of $2$.
\begin{claim*}
  Fix a choice $T$ of cells we wish to fill in,
  such that whenever a cell is in $T$,
  so are all the cells above and left of it.
  (In other words, $T$ is a Young tableau.)
  The number of ways to fill in these cells with sets
  satisfying the inclusion conditions is $2^{|T|}$.
\end{claim*}
An example is shown below, with an indeterminate set marked in red
(and the rest of $T$ marked in blue).
\[
  \begin{bmatrix}
    1234 & 12345 & 123456 & 1234567 & 12345678\\
    123 & {\color{blue}1234} & {\color{blue}12346} & {\color{blue}123467} \\
    12 & {\color{blue}124} & {\color{red}1234 \text{ or } 1246} \\
    1 & {\color{blue}12} \\
    \varnothing & {\color{blue}2}
  \end{bmatrix}
\]
\begin{proof}
  The proof is by induction on $|T|$,
  with $|T| = 0$ being vacuous.

  Now suppose we have a corner $\begin{bmatrix}
  B & C \\ A & {\color{red}S} \end{bmatrix}$
  where $A$, $B$, $C$ are fixed and $S$ is to be chosen.
  Then we may write $B = A \cup \{x\}$ and $C = A \cup \{x,y\}$
  for $x,y \notin A$.
  Then the two choices of $S$ are $A \cup \{x\}$ (i.e.\ $B$)
  and $A \cup \{y\}$, and both of them are seen to be valid.

  In this way, we gain a factor of $2$
  any time we add one cell as above to $T$.
  Since we can achieve any Young tableau in this way,
  the induction is complete.
\end{proof}