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 $i \neq j(i, j \in\{1,2, \cdots, k\}), A_{i} \cap A_{j}$ 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 the conditions: $A_{i}=\{2 n-i, 2 n\}(i=1,2, \cdots, n-1, n+1, \cdots, 2 n-1)$, $A_{n}=\{2 n\}$.
Second, we prove: $k \leqslant 2 n-1$.
If not, suppose $S$ has $2 n$ subsets $A_{1}, A_{2}, \cdots, A_{2 n}$ that simultaneously satisfy conditions (1) (2) (3). We call a pair of numbers $i, j$ that satisfy (3) partners, and use $|M|$ to denote the number of elements in set $M$.
First, we give a lemma (proof omitted).
Lemma In a graph with an odd number of vertices, there must be a vertex with an even degree.
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 partners of $b_{i}$ in $A_{1}$ is odd.
Let the points corresponding to $b_{1}, b_{2}, \cdots, b_{2 m-1}$ be $B_{1}, B_{2}, \cdots, B_{2 m-1}$.
If $b_{i}, b_{j} \in A_{1}$ are partners, then draw a line between the corresponding two points. In the graph formed by these points, each vertex has an odd degree, which is impossible by the lemma.
(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 $b_{i}$ except 1. Then $\left|C_{i}\right|$ is odd. Thus, $\sum_{i=1}^{2 m}\left|C_{i}\right|$ is even.

Now consider the $2 n-1$ numbers $2,3, \cdots, 2 n$, there must be one that appears in an even number of $C_{i}$ (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 partners of 1 and $t$ is even, i.e., $\left|A_{1} \cap A_{t}\right|$ is even, which contradicts the assumption.
In summary, $k \leqslant 2 n-1$.