For $n \in \mathbf{N}^{+}, A_{1}, A_{2}, \cdots, A_{2 n+1}$ are subsets of some set $B$. Suppose:
(1) $\left|A_{i}\right|=2 n(i=1,2, \cdots, 2 n+1)$;
(2) $\left|A_{i} \cap A_{j}\right|=1(1 \leqslant i<j \leqslant 2 n+1)$;
(3) Each element in $B$ belongs to at least two of the $A_i$'s.

For which $n$ can the elements of $B$ be labeled with 0 or 1 such that each $A_{i}$ has exactly $n$ elements labeled 0?
First, each element in $B$ belongs to exactly two of $A_{i}(i=1,2, \cdots, 2 n+1)$.
Otherwise, suppose $b \in B, b \in A_{1} \cap A_{2} \cap A_{3}$, then $A_{1} \cap A_{4}, A_{1} \cap A_{5}, \cdots, A_{1} \cap A_{2 n+1}$ can have at most $2 n-2$ distinct elements, and $A_{1}$ cannot contain any other elements, so $A_{1}$ can contain at most $2 n-1$ elements, which is a contradiction.

Second, construct a complete graph $k_{2 n+1}$ with $A_{1}, A_{2}, \cdots, A_{2 n+1}$ as vertices, and the edge $A_{i} A_{j}$ can be represented by the unique element $B_{i j}$ in $A_{i} \cap A_{j}$. Therefore, the elements of $B$ correspond one-to-one with the edges of $k_{2 n+1}$.
If $n$ is a positive integer that satisfies the problem's conditions, then $\frac{1}{2}|B|$ of the points in $B$ are labeled as 0.
Thus, $\frac{1}{2} \cdot(2 n+1) \cdot 2 n=n(2 n+1)$ is even, which implies that $n$ is even.
In fact, when $n=2 m$, such labeling can be achieved:
$b_{i j}$ is labeled as 0 if and only if $|i-j| \leqslant m(\bmod 2 n+1)$, and it is easy to verify that this labeling satisfies the problem's conditions.
!Comment First, combining condition (2), deduce from condition (3) that each element in $B$ belongs to exactly two of $A_{i}(i=1,2, \cdots, 2 n+1)$, then establish a one-to-one correspondence between the elements of $B$ and the edges of the complete graph $k_{2 n+1}$ with $A_{1}, A_{2}, \cdots, A_{2 n+1}$ as vertices, and finally consider labeling some points in $A_{i} \cap A_{j}$ as 0, thus solving the problem.