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

For which $n$ is it possible to label each element of $B$ with a 0 or 1 such that each $A$ contains exactly $n$ elements labeled 0?
First, the "at least" in (3) can actually be changed to "exactly". Because if there is an element $a_{1} \in A_{1} \cap A_{2 n} \cap A_{2 n+1}$, then the remaining $2 n-2$ subsets $A_{2}, A_{3}, \cdots, A_{2 n-1}$ each contain at most one element from $A_{1}$, thus there is at least one element in $A_{1}$ that does not belong to $A_{2} \cup A_{3} \cup \cdots \cup A_{2 n-1} \cup A_{2 n} \cup A_{2 n+1}$, which contradicts (3).

Therefore, construct a complete graph $K_{2 n+1}$, where each vertex $v_{i}$ represents a subset $A_{i}$, and each edge $\left(v_{i}, v_{j}\right) = b_{i j} (1 \leqslant i, j \leqslant 2 n+1, i \neq j)$ represents the element shared by sets $A_{i}$ and $A_{j}$. Thus, the problem is transformed into: for which $n$, can we label each edge of $K_{2 n+1}$ with either 0 or 1, such that from any vertex $v_{i}$, exactly $n$ of the $2 n$ edges are labeled with 0.

Since $K_{2 n+1}$ has $n(2 n+1)$ edges, if the above labeling requirement can be met, then the number of edges labeled with 0 is $\frac{1}{2} n(2 n+1)$, so $n$ must be even.

Conversely, if $n=2 m$ is even, we label the edges $\left(v_{i}, v_{i-m}\right), \left(v_{i}, v_{i-m+1}\right), \cdots, \left(v_{i}, v_{i-1}\right), \left(v_{i}, v_{i+1}\right), \cdots, \left(v_{i}, v_{i+m}\right), i=1,2, \cdots, 2 n+1$, all with 0, and the rest with 1, then we get the required labeling method (note that the addition of vertex indices is performed modulo $2 n+1$, i.e., $\left.v_{(2 n+1)+i}=v_{i}\right)$.
Therefore, the requirement can be met if and only if $n$ is even.