2. Let $n$ be a positive integer, and let $A_{1}, A_{2}, \cdots, A_{2n+1}$ be subsets of some set $B$. Suppose
(a) each $A_{i}$ contains $2 n$ elements,
(b) $A_{i} \cap A_{j} (1 \leqslant i<j \leqslant 2 n+1)$ contains exactly one element,
(c) every element of $B$ belongs to at least two subsets $A_{i}$.

Question: For which values of $n$ can a label of 0 or 1 be assigned to each element of $B$ such that each $A_{i}$ contains exactly $n$ elements labeled with 0?
2. We first prove that conditions $(a), (b), (c)$ together can imply a stronger condition than (c), namely
( $C^{*}$ ) Each element in $B$ belongs to exactly two subsets $A_{i}$. First, we have the equation
$$
\begin{array}{l} 
A_{1}=\left(A_{2} \cap A_{1}\right) \cup\left(A_{3} \cap A_{1}\right) \cdots \\
\cup\left(A_{2n+1} \cap A_{1}\right) .
\end{array}
$$

The set on the left side of the equation contains the right side, which is obvious. Condition (c) then shows that the right side of the equation also contains the left side. More generally, we have
$$
A_{1}=\bigcup_{\substack{i=1 \\ i \neq j}}^{2n+1}\left(A_{i} \cap A_{j}\right) \text {, }
$$

where $j=1,2, \cdots, 2n+1$.
If ( $C^{*}$ ) is not true, that is, there is an element $a$ in $B$ that belongs to at least three subsets, without loss of generality, let $a \in A_{1} \cap A_{2} \cap A_{3}$. At this point, due to (b),
$$
A_{1} \cap A_{2}=\{a\}, A_{1} \cap A_{3}=\{a\} .
$$

Thus, $\left(A_{1} \cap A_{2}\right) \cup\left(A_{1} \cap A_{3}\right)=\{a\}$. And for $j>3$, $A_{1} \cap A_{j}$ has only one element. From equation (1), it is clear that $A_{1}$ has at most $2n-1$ elements, which contradicts condition (a). This proves ( $C^{*}$ ).

This means that for a set $B$ with properties (a), (b), and (c), each element can correspond to a pair of natural numbers. Specifically, if $a \in B$ and $a \in A_{i} \cap A_{j}$, then let $a$ correspond to $(i, j)$, where $i \neq j$ and $i, j=1,2, \cdots, 2n+1$. Of course, $(i, j)$ and $(j, i)$ correspond to the same element.

Now, suppose we can label as required, then in the above pairs, exactly one class corresponds to 0, i.e.,
$$
\binom{2n+1}{2}=\frac{(2n+1)(2n)}{4}=\frac{n(2n+1)}{2}
$$

pairs can correspond, so the above fraction must be a natural number, thus $n$ must be even.

Conversely, if $n$ is even, we prove that a labeling method that meets the requirements exists. Divide a circle into $2n+1$ equal parts using $2n+1$ points. Place the numbers $1,2, \cdots, 2n+1$ on these points in sequence. For any two different numbers $i, j$ in $\{1,2, \cdots, 2n+1\}$, consider the shorter arc distance between them on the circle. If $i$ and $j$ are separated by an odd number of arcs, label the element in $A_{i} \cap A_{j}$ as 1; if they are separated by an even number of arcs, label it as 0. Since $n$ is even, half of the elements in $A_{1}$ are labeled as 0, and the other half are labeled as 1.