We know the Oddtown theorem: for any set $S$ with $|S|=n$ and $T \subseteq 2^{S}$ with each element of $T$ having odd cardinality and the intersection of two distinct elements of $T$ having even cardinality, $|T| \leq n$.

To show the following:

for any set $S$ with $|S|=n$ and $T \subseteq 2^{S}$ with each element of $T$ having even cardinality and the intersection of two distinct elements of $T$ having odd cardinality, $|T| \leq n+1$

we just add a new element to $S$ and all the elements of $T$, apply the Oddtown theorem, and then remove this element from $S$ and all the elemets of $T$.

We know the Oddtown theorem: for any set $S$ with $|S|=n$ and $T \subseteq 2^{S}$ with each element of $T$ having odd cardinality and the intersection of two distinct elements of $T$ having even cardinality, $|T| \leq n$.

To show the following:

for any set $S$ with $|S|=n$ and $T \subseteq 2^{S}$ with each element of $T$ having even cardinality and the intersection of two distinct elements of $T$ having odd cardinality, $|T| \leq n+1$

we just add a new element to $S$ and all the elements of $T$, apply the Oddtown theorem, and then remove this element from $S$ and all the elements of $T$.