C31 (29-2, Czechoslovakia) Let $n$ be a positive integer, and $A_{1}, A_{2}, \cdots, A_{2 n+1}$ be a family of subsets of a set $B$, satisfying the conditions:
(1) Each $A_{i}$ contains exactly $2 n$ elements;
(2) $A_{i} \cap A_{j}(1 \leqslant i<j \leqslant 2 n+1)$ contains exactly one element;
(3) Each element of $B$ belongs to at least two subsets $A_{i_{1}}$ and $A_{i_{2}}, 1 \leqslant i_{1}<i_{2} \leqslant 2 n$ +1.

For which positive integers $n$ is it possible to label each element of $B$ with a tag that has either 0 or 1 written on it, such that each $A_{i}$ contains exactly $n$ elements labeled with 0? Explain your reasoning.
Solution 1 When and only when $n$ is even, there is a way to label that meets the requirements.
First, we prove that from conditions (1), (2), and (3), we can derive a stronger conclusion than condition (3): every element in $B$ belongs to exactly two subsets $A_{i}(1 \leqslant i \leqslant 2 n+1)$.

In fact, if there is an element $a$ in $B$ that belongs to 3 or more subsets, without loss of generality, let $a \in A_{1} \cap A_{2} \cap A_{3}$. By condition (2), the remaining $2 n-2$ subsets $A_{4}, \cdots, A_{2 n+1}$ each contain at most one element from $A_{1}$, so there is at least one element in $A_{1}$ that does not belong to $A_{2} \cup A_{3} \cup \cdots \cup A_{2 n+1}$, which contradicts condition (3).

Thus, for a set $B$ that satisfies conditions (1), (2), and (3), each of its elements can correspond to a pair of positive integers, i.e., if $a \in B$ and $a \in A_{i} \cap A_{j}(1 \leqslant i<j \leqslant 2 n+1)$, then let $a$ correspond to the unordered pair $(i, j)$.

If we can label the elements with 0 or 1 as required, then in the above pairs, exactly half correspond to 0, so we have
$$
\frac{1}{2} \mathrm{C}_{2 n+1}^{2}=\frac{n(2 n+1)}{2}
$$

0s, thus $2 \mid n(2 n+1)$, so $n$ is even.
Furthermore, when $n$ is even, we can construct a labeling method that meets the requirements. Divide a circle into $2 n+1$ equal parts using $2 n+1$ points. Label these points with the numbers $1,2, \cdots, 2 n+1$ in sequence. For any two different numbers $i, j$ in $\{1,2, \cdots, 2 n+1\}$, consider the shorter arc between them on the circle. If $i$ and $j$ are separated by an odd number of arcs, label the element in $A_{i} \cap A_{j}$ with 1; if $i$ and $j$ are separated by an even number of arcs, label the element in $A_{i} \cap A_{j}$ with 0. Since $n$ is even, half of the elements in $A_{i}$ are labeled with 0, and the other half with 1.

Solution 2 From the given conditions, we know that each $A_{i}$ contains $2 n$ elements, which are exactly one common element with each of the other $2 n$ subsets, and $B=A_{1} \cup A_{2} \cup \cdots \cup A_{2 n+1}$. Therefore, the elements of $B$ can be arranged into a $2 n \times 2 n$ symmetric matrix $\left(a_{i j}\right)$, where $a_{i j}=a_{j i}=A_{i} \cap A_{j}$ $(1 \leqslant i<j \leqslant 2 n), a_{i j}=A_{i} \cap A_{2 n+1}(1 \leqslant i \leqslant 2 n)$, and the $i$-th row (or column) of the matrix $\left(a_{i j}\right)$ is $A_{i}$, and the main diagonal is $A_{2 n+1}$.

If we can label the elements with 0 or 1 as required, then the matrix $\left(a_{i j}\right)$ contains $2 n \cdot n=2 n^{2}$ 0s. Since it is a symmetric matrix, the number of 0s outside the main diagonal is even. Therefore, the number of 0s on the diagonal is also even, i.e., $n$ is even.
On the other hand, for $n=2 k$, we can arrange the $4 \times 4$ matrix:
$$
T=\left(\begin{array}{llll}
0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1
\end{array}\right)
$$

repeatedly to form a $2 n \times 2 n$ symmetric matrix:
$$
\left(\begin{array}{cccc}
T & T & \cdots & T \\
T & T & \cdots & T \\
\vdots & \vdots & \ddots & \vdots \\
T & T & \cdots & T
\end{array}\right)
$$

Label the elements of $B$ according to this matrix, and it will satisfy the requirements.