Is it possible to have more than $N = \binom{\lfloor n/2\rfloor}{2}$ subsets of an $n$-set, each of size 4, such that each two of them intersect in 0 or 2 elements?


To see that $N$ is achievable, choose $\lfloor n/2\rfloor$ disjoint pairs and then take each 4-set consisting of two of the pairs. But this is not the unique way of doing it in general.


EDIT: Patricia has provided a counterexample with $n=7$, so I'll remove odd $n$ from the question. Is there a counterexample for even $n$?



Intersecting 4-sets
The conjectured maximum of $N = \binom{\lfloor n/2\rfloor}{2}$
is correct except for $n=7$, when the maximum is $7$, and
$8 \leq n \leq 11$, when the maximum is $14$. The maximal
configuration is unique except for $n=12$, $13$, $15$, $16$, and $17$.


Let $L$ be the subgroup of ${\bf Z}^n$ generated by $(2{\bf Z})^n$
and the characteristic functions $e\_i + e\_j + e\_k + e\_l$ of each
4-set $\lbrace i,j,k,l \rbrace$ in our family $\cal F$ of subsets of
$\lbrace 1,2,\ldots,n \rbrace$. Give $L$ the structure of lattice
using the inner product 
 $$ \langle x, y \rangle = \frac12 \sum\_{i=1}^n x\_i y\_i $$
(i.e. *half* the usual inner product). Then $L$ is generated by
vectors $2e\_i$ and $e\_i + e\_j + e\_k + e\_l$ of norm $2$, any two of which
are either orthogonal or have inner product $1$. Hence $L$ is an
even integral lattice, with at least $2n+16|{\cal F}|$ roots
(vectors of norm 2), namely $\pm 2 e\_i$ and $\pm e\_i \pm e\_j \pm e\_k \pm e\_l$
for $\lbrace i,j,k,l \rbrace \in \cal F$. Equality holds iff $\cal F$ contains every tetrad $\lbrace i,j,k,l \rbrace$ such that $e\_i + e\_j + e\_k + e\_l \in L$.


Now we can use the theory of *root systems* to partition the set of
roots of $L$ into mutually orthogonal simple root systems. Since
$L$ contains the root lattice $A\_1^n = (2{\bf Z})^n$, the only possible
components of the root system of $L$ are $A\_1$, $D\_{2k}$ for $k \geq 2$,
and the exceptional systems $E\_7$ and $E\_8$. These contribute respectively
$0$, $\binom{k}{2}$, $7$ and $14$ tetrads to $\cal F$. Namely, each $A\_1$
corresponds to a coordinate that does not appear in $\cal F$; each $D\_{2k}$
corresponds to $k$ pairs of coordinates paired in each of $\binom{k}{2}$
possible ways; and $E\_7$ and $E\_8$ correspond to the tetrads of the
Hamming $[7,3,4]$ and extended Hamming $[8,4,4]$ codes respectively.


It is now elementary bookkeeping to obtain the maximum configuration.


$\circ$ Except for $7 \leq n \leq 11$, the maximal $|{\cal F}|$ is $\binom{k}{2}$
for $n = 2k$ or $n = 2k+1$, attained by the $D\_{2k}$ configuration.


$\circ$ For $n=7$, the maximum of $7$ is attained by the $E\_7$ (Hamming) configuration,
and for $8 \leq n \leq 11$, by $E\_8 \oplus A\_1^{n-8}$ (extended Hamming).


$\circ$ For $n=12$ ($n=13$), the maximum of $15$ is attained by both
$D\_{12}$ ($D\_{12} \oplus A\_1$) and $E\_8 \oplus D\_4$ ($E\_8 \oplus D\_4 \oplus A\_1$).


$\circ$ For $n=15$, the maximum of $21$ is attained by both
$D\_{14} \oplus A\_1$ and $E\_8 \oplus E\_7$.


$\circ$ Finally, for $n=16$ ($n=17$),
the maximum of $28$ is attained by both
$D\_{16}$ ($D\_{16} \oplus A\_1$) and
$E\_8 \oplus E\_8$ ($E\_8 \oplus E\_8 \oplus A\_1$).


[The lattice $L$ corresponds via "construction A" to a binary linear code
generated by $\cal F$, which is doubly even by hypothesis. Koch developed
a theory of "[tetrad systems](http://www.mathunion.org/ICM/ICM1986.1/Main/icm1986.1.0457.0465.ocr.pdf)" of such codes that could be used to
give a more direct but less familiar derivation of this answer.]


