There is an example for $n=6$, a quite exceptional one. It comes from the smallest finite sporadic simple group $G=M_{11}$ of order 7920. $G$ has a permutation representation on 12 points, corresponding to the (left) cosets of a subgroup isomorphic to $PSL_2(11)$. A subgroup isomorphic to $A_6$, the alternating group of degree 6, has 2 orbits on 12 points, both of length 6, and the images of these orbits under $G$ are the 22 blocks of the design.

In general, consider the $(2n\times 2n)$-matrix $H$ with entries $\pm 1$ columns labelled by points, the 1st row of all 1s, and the remaining rows labelled by pairs $b,b'$ of complementary blocks (we can assume w.l.o.g. that $b$ contains the 1st point); we put 1 in the $(b,b'),p$-entry of $H$ if the point $p$ is in $b$, and -1 otherwise. If $q$ is another point then, as the pair $(p,q)$ lies in $n-1$ blocks, the scalar product of $p$th and $q$th columns of $H$ is $1+(n-1)-n=0$. Thus $H$ is a Hadamard matrix. As Chris points out, this also means that the design is a 3-design, known as Hadamard 3-design. (It will be a $3-(2n,n,n/2-1)$-design).

Thus, indeed, we will always have a Hadamard matrix giving us the design in question; in particular $n$ must be even.