[proofplan] The necessity of the admissibility congruences is a double-counting argument: first count pairs containing a fixed point, and then count all ordered pairs of distinct points. The sufficiency direction is the hard asymptotic existence input: we cite Wilson's original PBD existence theorem as an external theorem and apply it to the nonempty finite block-size set $K$ after verifying the two divisibility hypotheses in the definition of admissibility. Combining these two directions gives the claimed eventual if and only if statement. [/proofplan] [step:Derive the pointwise divisibility condition from blocks through one point] Assume that a $\operatorname{PBD}(v,K,1)$ exists. Let $X$ be its point set, so $|X|=v$, and let $\mathcal B$ be its set of blocks. For a point $x\in X$, define \begin{align*} \mathcal B_x:=\{B\in\mathcal B:x\in B\}. \end{align*} Because every unordered pair of distinct points of $X$ lies in exactly one block, the sets $B\setminus\{x\}$ with $B\in\mathcal B_x$ form a partition of $X\setminus\{x\}$. Hence \begin{align*} v-1=\sum_{B\in\mathcal B_x}(|B|-1). \end{align*} For every $B\in\mathcal B_x$, the block size $|B|$ belongs to $K$, so $\alpha(K)=\gcd\{k-1:k\in K\}$ divides $|B|-1$. Therefore $\alpha(K)$ divides the sum, and $\alpha(K)\mid v-1$. [/step] [step:Derive the global divisibility condition by counting ordered pairs] Again assume that $(X,\mathcal B)$ is a $\operatorname{PBD}(v,K,1)$. Count ordered pairs $(x,y)\in X\times X$ with $x\ne y$. There are $v(v-1)$ such pairs. On the other hand, a block $B\in\mathcal B$ contributes exactly $|B|(|B|-1)$ ordered pairs with both entries in $B$, and the pairwise balanced condition says that each ordered pair of distinct points is contributed by exactly one block. Thus \begin{align*} v(v-1)=\sum_{B\in\mathcal B}|B|(|B|-1). \end{align*} For every $B\in\mathcal B$, the integer $|B|$ lies in $K$, so $\beta(K)=\gcd\{k(k-1):k\in K\}$ divides $|B|(|B|-1)$. Hence $\beta(K)$ divides the displayed sum, and $\beta(K)\mid v(v-1)$. [guided] We now count the same finite set in two ways. Let \begin{align*} P:=\{(x,y)\in X\times X:x\ne y\} \end{align*} be the set of ordered pairs of distinct points. Directly, $|P|=v(v-1)$, since there are $v$ choices for $x$ and then $v-1$ choices for $y$. The block decomposition gives a second count. For each block $B\in\mathcal B$, define \begin{align*} P_B:=\{(x,y)\in B\times B:x\ne y\}. \end{align*} Then $|P_B|=|B|(|B|-1)$. The defining property of a $\operatorname{PBD}(v,K,1)$ says that every unordered pair $\{x,y\}$ with $x\ne y$ is contained in exactly one block. Therefore the ordered pair $(x,y)$ is also contained in exactly one block, namely the unique block containing $\{x,y\}$. Hence the sets $P_B$, as $B$ ranges over $\mathcal B$, form a partition of $P$. Additivity of cardinality over a finite partition gives \begin{align*} v(v-1)=|P|=\sum_{B\in\mathcal B}|P_B|=\sum_{B\in\mathcal B}|B|(|B|-1). \end{align*} Since every block size $|B|$ belongs to $K$, each summand $|B|(|B|-1)$ is divisible by $\beta(K)=\gcd\{k(k-1):k\in K\}$. A common divisor of all summands divides their sum, so $\beta(K)\mid v(v-1)$. This is the second admissibility condition. [/guided] [/step] [step:Apply Wilson's external asymptotic sufficiency theorem to admissible orders] Let $K\subset\mathbb N$ be nonempty and finite with every $k\in K$ satisfying $k\ge 2$. The nonemptiness ensures that \begin{align*} \alpha(K)=\gcd\{k-1:k\in K\}\quad\text{and}\quad \beta(K)=\gcd\{k(k-1):k\in K\} \end{align*} are defined positive integers. We invoke Wilson's external PBD existence theorem, proved in R. M. Wilson, ["An existence theorem for pairwise balanced designs, III: proof of the existence conjecture"](https://doi.org/10.1016/0097-3165(75)90067-9), Journal of Combinatorial Theory, Series A 18 (1975), 71-79. In the form needed here, Wilson's theorem states that for every nonempty finite set $K$ of integers at least $2$ there is an integer $N(K)$ such that whenever an integer $v\ge N(K)$ satisfies \begin{align*} \alpha(K)\mid v-1 \quad\text{and}\quad \beta(K)\mid v(v-1), \end{align*} there exists a $\operatorname{PBD}(v,K,1)$. The hypotheses of that theorem are met because $K$ is nonempty, finite, and all its elements are at least $2$. If $v$ is admissible for $K$, the two displayed divisibility conditions hold by definition, so Wilson's external theorem produces a $\operatorname{PBD}(v,K,1)$ for every $v\ge N(K)$. [/step] [step:Choose the threshold and combine the two implications] Define $v_0:=N(K)$, where $N(K)$ is the threshold supplied by Wilson's external asymptotic sufficiency theorem. For every integer $v\ge v_0$, the first two steps show that the existence of a $\operatorname{PBD}(v,K,1)$ implies that $v$ is admissible for $K$. The preceding step shows that admissibility of $v$ implies the existence of a $\operatorname{PBD}(v,K,1)$. Therefore, for every $v\ge v_0$, a $\operatorname{PBD}(v,K,1)$ exists if and only if $v$ is admissible for $K$. [/step]