4. Let $n$ be a non-negative integer. Find the number of ways to select the family of sets $S_{i, j} \subseteq\{1,2, \cdots, 2 n\}(0 \leqslant i \leqslant n, 0 \leqslant j \leqslant n$, which can be the same) that satisfy the following conditions:
(1) $\left|S_{i, j}\right|=i+j$;
(2) When $0 \leqslant i \leqslant k \leqslant n$ and $0 \leqslant j \leqslant l \leqslant n$, we have $S_{i, j} \subseteq S_{k, l}$.
4. Note that,
$$
\begin{aligned}
\varnothing & =S_{0,0} \varsubsetneqq S_{0,1} \varsubsetneqq \cdots \varsubsetneqq S_{0, n} \varsubsetneqq S_{1, n} \varsubsetneqq \cdots \varsubsetneqq S_{n, n} \\
= & \{1, \cdots, 2 n\} .
\end{aligned}
$$

From this, we know that the selection method for the sets $S_{0,0}, S_{0,1}, \cdots, S_{0, n}, S_{1, n}, \cdots, S_{n, n}$ has a total of $(2 n)!$ ways.

Assuming these $2 n+1$ sets have been determined in any one of these ways, consider the square matrix:
$$
\left(\begin{array}{cccc}
S_{0, n} & S_{1, n} & \cdots & S_{n, n} \\
\vdots & \vdots & \ddots & \vdots \\
S_{0,1} & S_{1,1} & \cdots & S_{n, 1} \\
S_{0,0} & S_{1,0} & \cdots & S_{n, 0}
\end{array}\right),
$$

where the sets in the first row and the first column are already determined, and satisfy equation (1).

The problem conditions imply that the matrix should satisfy for any $i, j$, $S_{i, j} \subseteq S_{i+1, j}, S_{i, j} \subseteq S_{i, j+1}$, and $S_{i+1, j}$ and $S_{i, j+1}$ each contain exactly one element not in $S_{i, j}$.

For any $2 \times 2$ submatrix $\left(\begin{array}{ll}B & C \\ A & S\end{array}\right)$, if the sets $A, B, C$ are already determined, then there exist $x, y$ such that $x, y \notin A$, and $B=A \cup\{x\}, C=A \cup\{x, y\}$. Thus, the set $S$ has exactly two valid selection methods:
$$
S=A \cup\{x\} \text { or } S=A \cup\{y\} \text {. }
$$

Therefore, when the first row and the first column of the entire matrix are already determined, $S_{1, n-1}$ has two selection methods; when $S_{1, n-1}$ is also determined, $S_{2, n-1}$ and $S_{1, n-2}$ each have two selection methods (and are independent of each other); $\cdots \cdots$ when $S_{n, 1}$ and $S_{n-1,0}$ are also determined, $S_{n, 0}$ has two selection methods. Thus, by the multiplication principle, we know that the number of ways to select the set family $S_{i, j}(0 \leqslant i \leqslant n, 0 \leqslant j \leqslant n)$ that meets the requirements is $(2 n)!\times 2^{n^{2}}$.