Let $ n$ be an even positive integer. Let $ A_1, A_2, \ldots, A_{n \plus{} 1}$ be sets having $ n$ elements each such that any two of them have exactly one element in common while every element of their union belongs to at least two of the given sets. For which $ n$ can one assign to every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly $ \frac {n}{2}$ zeros?
1. **Claim 1: Each element of the union belongs to exactly 2 subsets.**

   Proof:
   Consider a subset \( A_i \). Assume that some element \( x \in A_i \) also belongs to \( A_k \) and \( A_l \). There are \( n-1 \) elements remaining in \( A_i \) and there are \( n-2 \) subsets to choose from. By the pigeonhole principle, at least one of the remaining elements in \( A_i \) must belong to \( A_k \) or \( A_l \). This contradicts the assumption that any two subsets have only one element in common. Therefore, each element of the union belongs to exactly 2 subsets.

2. **Claim 2: \( 4 \mid n \)**

   Proof:
   Since each element in \( A_i \) belongs to exactly one other subset, the total number of elements present in the union is \( \frac{n(n+1)}{2} \).

   If each subset must have \( \frac{n}{2} \) elements assigned a value of 1, the total number of elements assigned the value of 1 is:
   \[
   \frac{n}{2} \times \frac{n+1}{2} = \frac{n(n+1)}{4}
   \]
   Thus, 4 must divide \( n \).

3. **Claim 3: \( 4 \mid n \) is a sufficient condition to assign every element of the union one of the numbers 0 and 1 in such a manner that each of the sets has exactly \( \frac{n}{2} \) zeros.**

   Proof:
   Consider a regular polygon consisting of \( n+1 \) vertices where each line joining two vertices \( A_i, A_j \) represents the element which belongs to \( A_i \) and \( A_j \). Clearly, there are a total of \( \frac{n(n+1)}{2} \) such lines representing the total number of elements of the union where each vertex is connected to \( n \) vertices, meaning each of the \( n \) elements of \( A_i \) is part of one other subset.

   Starting with \( i = 1 \), let us start coloring all lines joining vertices \( A_i, A_{i+1} \) with color Red, all lines joining \( A_i, A_{i+2} \) with color White, \( A_i, A_{i+3} \) with color Red, \( A_i, A_{i+4} \) with color White, and so on until \( A_i, A_{i+n/2} \) with color White.

   Clearly, each line from vertex \( A_i \) alternates Red, White for the first \( n/2 \) lines and then alternates White, Red for the remaining \( n/2 \) lines, implying that we could have exactly \( n/2 \) red lines emanating from each vertex \( A_i \). But these \( n/2 \) lines represent \( n/2 \) elements of each subset \( A_i \) which could each be assigned a value of 0. This completes the proof.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) such that \( 4 \mid n \).