The answer is indeed asymptotic to $4n/\log(n)$, but I don't know of an elementary or easy construction for this upper bound. I believe simple probabilistic-method type constructions do not work.
This is also phrased as a "coin-weighing" problem. One point to keep in mind is that we can think of each weighing of $k$ balls as simply learning how many of the $k$ are light versus heavy. Thus you can rephrase your problem as follows: There is a special unknown subset $S$ of $\{1,\dots,2n\}$, where $|S| = n$, and we wish to determine $S$. At each step, we can query any $T \subseteq \{1,\dots,2n\}$ and learn $|S \cap T|$.
I find these references in the tutorial/lecture by Galvin[1], where he attributes the upper bound to (independently) [2,3] and the lower bound to (independently) [4,5]. Note that in his formulation there are only $n$ coins rather than $2n$ (hence the bound is $2n/\log n$), and it is not promised how many are heavy or light (so in particular the upper bound holds for the case where half are promised to be heavy).
This is also addressed in this MO question[6], but I'm not sure where a factor $2$ has gone missing in that answer.
[1] http://arxiv.org/abs/1406.7872
[2] D. Cantor and W. Mills, Determination of a subset from certain combinatorial properties, Can. J. Math. 18 (1966), 42–48
[3] B. Lindstr¨om, On a combinatorial problem in number theory, Can. Math. Bull. 8 (1965), 477-490.
[4] P. Erd˝os and A. R´enyi, On two problems of information theory, Publ. Hung. Acad. Sci. 8 (1963), 241–254.
[5] L. Moser, The second moment method in combinatorial analysis, in “Combinatorial Structures and Their Applications”, 283–384, Gordon and Breach, New York, 1970.
[6] Guessing a subset of {1,...,N}Guessing a subset of {1,...,N}
