First of all, the incidence structure is determined uniquely by the given conditions: every element of $A=\bigcup A_i$ belongs to precisely two $A_i$'s, and $|A|=\frac{n(n+1)}2$. We can see this as follows: Suppose we can find $x\in A_1\cap A_2\cap A_3$. Then $A_1\cup A_2\cup A_3$ has, except for $x$, at least $3(n-1)$ elements. Every $A_i,\ i>3$ covers at most $3$ of these (one from each of $A_1,A_2,A_3$), so in order to cover these elements one more time we would need at least $n-1$ more $A_i$'s, but we don't have that many. From this, the fact that $A$ has $\binom{n+1}2$ elements follows by a simple counting argument. Now let's observe that $n$ being a multiple of $4$ is a necessary condition for the existence of the $0-1$ labeling. This is beause if we count the $0$-labeled elements every time they appear in a set we get $\binom{n+1}2$ (since there are $n+1$ sets, each one containing $\frac n2$ of them), and an even number, since every $0$-labeled element appears in precisely two sets, so it's counted twice. This shows that $\binom{n+1}2$ must be even, which, in turn, means that $4|n$. Conversely, there is such a $0-1$ labeling when $4|n$, and we can prove it inductively. First of all, let's construct an $(n+1)\times n$ table $T_n$ containing numbers between $1$ and $\binom{n+1}2$ as follows: The rows of the table will be the sets $A_i$ in order, and the numbers in a row will be the elements in the respective set. The first row is just $1,2,\ldots,n$. In the $i$'th step (with $i\in\overline{2,n+1}$) we construct the $i$'th row by setting $a_{i,j}=a_{j,i-1}$ when $j\in\overline{1,i-1}$, and making $a_{i,i},a_{i,i+1},\ldots,a_{i,n}$ the smallest consecutive numbers larger than all the numbers which have appeared so far. The subtable formed by the last $n+1-4$ rows and $n-4$ columns has the same structure as $T_{n-4}$ (i.e. we can construct an isomorphism from $T_{n-4}$ to it induced by a bijection between the numbers $1,2,\ldots,\binom{n-4+1}2$ and those which appear in the mentioned subtable), so, by the induction hypothesis, we can find the desired $0-1$ labeling of this subtable. We divide the rest in three other subtables: $X$, the subtable formed by the first $4$ lines and $3$ columns, which we color $\begin{pmatrix}1&1&0\\1&0&1\\1&0&0\\0&1&0\end{pmatrix}$, $Y$, the one formed by the first $4$ lines and $n-3$ columns, whose columns we color alternately $\begin{pmatrix}0\\0\\1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\1\\0\\0\end{pmatrix}$ (starting with the first one), and the rest of the initial table, which lies in the lower-left corner, and is just the transpose of $Y$ (so its coloration is determined). Actually drawing these tables should help clarify the whole thing.

another way to see that there are $\frac{n(n+1)}{2}$ elements in $X=\bigcup_i A_i$ is to count the number of pairs $(x,i)$ where $x\in X,x\in A_i$; that counts that $n(n+1)\geq 2|X|$; on the other hand counting the number of triples $(i,j,x),x\in X,x\in A_i\cap A_j$ gives that $\sum_{x\in X} d_x^2=2n(n+1)$ where $d_x$ is the degree of each $x$(i.e. the number of sets that contains $x$). Use Cauchy and it follows that $2|X|\geq n(n+1)$. in fact, if you observe, the set collection here is unique upto isomorphism(relabelling of the points).

Another way to look that 4/n is sufficient is to think the n+1 sets as the points of a graph, and their members as edges. So we need a n/2 regular graph on n+1 vertices which is easily constructed from the following general fact, Kn is the disjoint union of hamilton circuit iff n is odd. I didnt read Grobber post carefully but its probably constructing this.

Answer: All $n$ such that $4|n$ We first make the following $2$ claims: Claim $1$: Each element of union belongs to exactly $2$ subsets. Proof: Consider a subset $A_i$. Assume that some element $x \in\ A_i$ also $\in A_k, A_l$. There are $n-1$ elements remaining in $A_i$ and there are $n-2$ subsets to choose from. By pigeon hole principle, at least $1$ of the remaining elements in $A_i$ must $\in A_k$ or $\in A_l$. This contradicts the assumption that any $2$ subsets have only $1$ element in common. Claim $2$: $4|n$ Proof: Now, since each element in $A_i \in$ exactly $1$ other subset, total number of elements present in the union = $n*(n+1)/2$. If each subset must have $n/2$ elements assigned a value of $1$, the total number of elements assigned value of $1$ = $n/2*(n+1)/2 = n*(n+1)/4$. Thus $4$ must divide $n$. Now we make our final claim: Claim $3$: $4|n$ 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 $\in A_i, A_j$. Clearly there are a total of $n*(n+1)/2$ 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 $1$ 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 ... $A_i$, $A_{i+n/2}$ with color White. Clearly each line from vertex $A_i$ alternates Red, White for first $n/2$ lines and then alternates White, Red for 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.