Four. (50 points) Let $n \in \mathbf{N}_{+}, S=\{1,2, \cdots, 2 n\}$ have $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$ satisfying:
(1) For any $i \neq j(i, j=1,2, \cdots, k)$, $A_{i} \cap A_{j}$ has exactly an odd number of elements;
(2) For any $i(i=1,2, \cdots, k)$, $i \notin A_{i}$;
(3) If $i \in A_{j}$, then $j \in A_{i}$.
Determine the maximum value of $k$.
Four, $k_{\max }=2 n-1$.
First, the following $2 n-1$ sets satisfy conditions (1), (2), and (3):
$$
\begin{aligned}
A_{i} & =\{2 n-i, 2 n\}(i=1,2, \cdots, n-1, n+1, \cdots, 2 n-1), \\
A_{n} & =\{2 n\} .
\end{aligned}
$$

Next, we prove: $k \leqslant 2 n-1$.
If not, suppose the $2 n$ subsets $A_{1}, A_{2}, \cdots, A_{2 n}$ of $S$ simultaneously satisfy (1), (2), and (3).

We call a pair of numbers $i, j(i \neq j)$ that satisfy (3) a "connection". Let $|M|$ denote the number of elements in set $M$.
First, we give a lemma.
Lemma: In a graph with an odd number of vertices, there must be a vertex with an even degree.
Proof is omitted.
Back to the original problem.
(1) If there exists $i(1 \leqslant i \leqslant 2 n)$ such that $\left|A_{i}\right|$ is odd, without loss of generality, let $A_{1}=\left\{b_{1}, b_{2}, \cdots, b_{2 m-1}\right\}$. Then for each $b_{i}(1 \leqslant i \leqslant 2 m-1)$, by the problem's condition, the number of connections of $b_{i}$ in $A_{1}$ is odd.

Let $b_{1}, b_{2}, \cdots, b_{2 m-1}$ correspond to the sets $B_{1}$, $B_{2}$, \cdots, $B_{2 m-1}$, respectively.

If $b_{i}$ and $b_{j}\left(\in A_{1}\right)$ are in a connection relationship, then connect a line between the corresponding two points. In the graph formed by these points, each vertex has an odd degree. By the lemma, this is impossible.
(2) If for any $i(i=1,2, \cdots, 2 n)$, $\left|A_{i}\right|$ is even, let $A_{1}=\left\{b_{1}, b_{2}, \cdots, b_{2 m}\right\}$. Let $C_{i}$ $(i=1,2, \cdots, 2 m)$ be the set of partners of $b_{i}$ excluding 1. Then $\left|C_{i}\right|$ is even. Thus, $\sum_{i=1}^{2 m}\left|C_{i}\right|$ is even.

Now consider the $2 n-1$ numbers $2,3, \cdots, 2 n$. Among them, there must be a number that appears in an even number of $C$'s (otherwise, the sum of an odd number of odd numbers is odd, i.e., the total number of appearances is odd, which contradicts $\sum_{i=1}^{2 m}\left|C_{i}\right|$ being even) (let this number be $t$), then the number of common connections between $1$ and $t$ is even, i.e., $\left|A_{1} \cap A_{t}\right|$ is even, which contradicts the assumption.
In summary, $k \leqslant 2 n-1$.
(Zhang Limin, Daqing Foreign Languages School, 163458)