I guess I will go through the solution as I found it. First of all, note that if your first measurement is imbalanced, you are probably home free. If the left is heavier than the right, then the right cannot have more than 1 fake mass, and therefore you can divide the right into two piles (throwing out an odd mass if you need to) and use those for your second weighing. Whichever side is lighter cannot have any fakes, and if they are both equal then neither has fakes. Thus if the first measurement has N masses to a side, then if it is unbalanced you can find N/2 genuine masses (or (N-1)/2 if N is odd).
This means that we only need to solve the case where the first measurement is balanced, and this is much more difficult to deal with (as you almost certainly found if you tried to solve it). My first solution found at least 10 genuine masses as follows: Begin by dividing up the masses into 4 groups, A has 10 masses, B has 20, C has 30 and D has 40. Next, weigh A+B against C, if this is imbalanced, use the method above to find at least 15 genuine masses. If it is balanced, this means that A+B and C each have the same number of fakes, either 0,1, or 2 each. This means that D must have 0,2, or 4 fakes, this restriction is very important. Next, weigh A+C against D. Suppose this is balanced as well. Then D must have 2 fakes (it can't have 0, for C would have 2 then), but if D has 2 fakes and this measurement is balanced, then B is free of fakes. Alright, suppose this measurement has D heavy, then if D has 2 fakes, C must have 1 and A+B has 1, but its not in A, so A is genuine, if D instead had 4 fakes, A is still genuine. Finally, if A+C is heavy, then D has 0 fakes.
So, each result in our second measurement points to one coin pile, D heavy implies A, balanced implies B and A+C heavy implies D. Thus we are guaranteed to find at least 10 genuine masses.
We can generalize this, have 4 yet unspecified piles, A,B,C,D and use the method above. This means we must obey the equations:
A+B=C
A+C=D
A+B+C+D=100
Three equations for 4 unknowns, so there is a one-dimensional degree of freedom. We use that freedom to maximize min{A,B,D}. To be strict we also want to make sure that (A+B)/2 isn't too small, just in case the first measurement is balanced, but that isn't really an issue.
One can solve those equations as to get
A=3D/2-50
B=100-2D
C=50-D/2
Just looking at integer values of D and maximizing min{A,B,D} the solution is when D=42
A=13 B=16,C=29
So this guarantees getting 13 genuine masses. One funny point, there is a "better" solution at D=300/7=42+6/7 with
A=100/7=14+2/7
B=100/7=14+2/7
C=200/7=28+4/7
So, if you can use fractional masses, this is better.
Anyway, on Tanya Khovanova’s Math Blog they discuss a solution with 14 masses, but never explain it fully. This might be related to my fractional solution, but that is slightly crazy. Anyway, this solution can be constructed using the following structure for piles: A=14, B=14, C=29, D=42, E=1
First weigh A+B+E vs C, if it is imbalanced, then you can find 14 genuine ones. If it is balanced, then D again has 0,2, or 4 fakes in it, corresponding to C having 2,1, and 0 fakes in it.
Next weigh A+C vs D+E. If this is balanced, then D must have 2 fakes, C has 1 and A has 1, so B+E are all genuine (had D no fakes, C would have 2 and E cannot make up for it). If this measurement has D+E heavy, then D cannot have 0 fakes (C would have 2) so if D has 2 fakes, C has 1 and A cannot have any, so A is genuine (if D has 4 fakes, A is also genuine). Finally, if our measurement has A+C heavy, then D cannot have 2 fakes in it, so D is genuine.
So we have A+C heavy implies D is genuine, D+E heavy implies A genuine, and balanced implies B+E genuine. This guarantees at least 14 genuine masses are found. The word genuine has lost all meaning to me at this point.
