Solution:

If for every $s=1,2, \ldots, B$ any $s$ boys know together at least $s$ girls then the Hall (marriages') theorem implies that every boy can dance with a known girl and the condition is satisfied.

Let us assume now the converse and choose the largest $s \leq B$, such that there are $s$ boys who know together at most $s-1$ girls.

Denote the set of these $s$ boys by $S$ and let $L$ be the set of girls known to the boys from $S$. If some $t$ of the boys outside $S$ know together at most $t$ of the girls outside $L$ we have a contradiction with the choice of $s$. Therefore every $t$ boys outside $S$ know together at least $t+1$ of the girls outside $L$.

Now the Hall theorem implies that every boy outside $S$ can dance with a known girl outside $L$. Hence still non-dancing girls outside $L$ are at least
$$G-(B-s)-(s-1)=G+1-B \geq B$$
(the girls which dance with boys outside $S$ are $B-s$ and the girls which are known to the boys from $S$ are at most $s-1$ ). If the boys from $S$ dance with some $s$ of these remaining non-dancing girls outside $L$ then the condition is satisfied.