Let $B$ denote the set of boys and let $G$ denote the set of girls.

Let $R\subseteq B\times G$ denote the (symmetric) relation of knowing
eachother.

For $g\in G$ let $B_{g}=\left\{ b\in B\mid\neg bRg\right\} $ and
for $K\subseteq G$ let $B_{K}=\bigcap_{g\in K}B_{g}$.

Then a group of boys $H\subseteq B$ is bad if and only if $H\in\wp\left(B_{g}\right)$
for some $g\in G$ and with inclusion/exclusion we find the cardinality
of the collection of these groups as the following expression:

$$\left|\bigcup_{g\in G}\wp\left(B_{g}\right)\right|=\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\cap_{g\in K}\wp\left(B_{g}\right)\right|=\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\wp\left(\cap_{g\in K}B_{g}\right)\right|=$$$$\sum_{\varnothing\neq K\subseteq G}\left(-1\right)^{^{\left|K\right|-1}}\left|\wp\left(B_{K}\right)\right|\tag1$$

We are not interested in the cardinality itself but in the parity
of this cardinality.

Further this parity is the same as the parity
of the cardinality of the collection of good groups of boys because
addition of both cardinalities must give the even number $2^{\left|B\right|}$.

Calculating modulo $2$ we find that only the terms in $(1)$ where $B_{K}=\varnothing$
matter, because the others are all even numbers. Also the powers of $-1$ are irrelevant by calculation modulo $2$.

This leads to the conclusion that the parity of the cardinality of the set $\left\{ K\in\wp\left(G\right)-\left\{ \varnothing\right\} \mid B_{K}=\varnothing\right\} $
must be the same.

Now observe that:

$$B_{K}=\varnothing\iff\neg\exists b\in B\forall g\in K\;\neg bRg\iff\forall b\in B\exists g\in K\;bRg\iff K\text{ is a good girl group}$$
This completes the proof.