CLAIM: If we know that a group of n coins has an odd number of reals, in one weighing we can identify a group of less than or equal to (n/2) + 1 coins (I know that's not always an integer, hence less than or equal to), that also contains an odd number of reals.
PROOF: Let 2x be the largest even number less than or equal to n/2. Put x coins on either side of the scale.
CASE 1: The scale reads an even amount. In this case, we KNOW that out of our 2x coins, we have an even number of fakes. This is easy to show by perturbation - imagine starting with x real coins on each side, and changing a coin to a fake requires changing another coin to a fake to keep the scale reading even. So if there are an even number of odds in our 2x coins, there are an even number of reals. This means there are an ODD number of reals in the coins we didn't weigh, which is at most (n/2)+1.
CASE 2: The scale reads an odd amount. For similar reasoning to above, we have an ODD number of fakes in our 2x coins, and an ODD number of reals in our 2x coins. This amount is less than or equal to n/2.

So you can see that we can identify a MUCH smaller sample space with an odd number of real coins.

Even cooler to note is that this algorithm works with coin roles reversed.

If the number of coins with an odd number of reals, n, is 0 mod 4, then x will be n/4. So our weighing, without a doubt, will lead to identifying a group of n/2 coins with an odd number of reals.
If the number of coins with an odd number of reals, n, is 1 mod 2 (i.e. 1 or 3 mod 4), then we KNOW we can split it into two groups that differ in size by only 1. One of these groups is even, in which case we set that to be 2x. So, if n is 1 mod 4, we know our weighing will lead to EITHER (n+1)/2 or (n-1)/2.
Last case is n is 2 mod 4. n = 4z + 2. Set x to be z. So we will either identify a group of size 2z, or 2z + 2, i.e. (n/2)+1 or (n/2)-1.

SIXTH WEIGH: Put ALL THE COINS on one side of the scale. If a real coin weighs w grams, then we will get a number 99w +- (some even number). We know the disposition of the fake coins is even, since there's an even number of fakes! So, we now KNOW the decimal residue of 99w mod 2, call it r.
99w mod 2 = r.
99(w+-1) mod 2 = r +- 99.
99 is odd, so a 99 fake coins would have a different mod residue to r, which we have already identified from our weighing.

CLAIM: If we know that a group of $n$ coins has an odd number of reals, in one weighing we can identify a group of less than or equal to $(n/2)+1$ coins (I know that's not always an integer, hence less than or equal to), that also contains an odd number of reals.
PROOF: Let $2x$ be the largest even number less than or equal to $n/2$. Put $x$ coins on either side of the scale.
CASE 1: The scale reads an even amount. In this case, we KNOW that out of our $2x$ coins, we have an even number of fakes. This is easy to show by perturbation - imagine starting with $x$ real coins on each side, and changing a coin to a fake requires changing another coin to a fake to keep the scale reading even. So if there are an even number of odds in our $2x$ coins, there are an even number of reals. This means there are an ODD number of reals in the coins we didn't weigh, which is at most $(n/2)+1$.
CASE 2: The scale reads an odd amount. For similar reasoning to above, we have an ODD number of fakes in our $2x$ coins, and an ODD number of reals in our $2x$ coins. This amount is less than or equal to $n/2$.

So you can see that we can identify a MUCH smaller sample space with an odd number of real coins.

Even cooler to note is that this algorithm works with coin roles reversed.

If the number of coins with an odd number of reals, $n$, is 0 mod 4, then $x$ will be $n/4$. So our weighing, without a doubt, will lead to identifying a group of $n/2$ coins with an odd number of reals.
If the number of coins with an odd number of reals, $n$, is 1 mod 2 (i.e. 1 or 3 mod 4), then we KNOW we can split it into two groups that differ in size by only 1. One of these groups is even, in which case we set that to be $2x$. So, if $n$ is 1 mod 4, we know our weighing will lead to EITHER $(n+1)/2$ or $(n-1)/2$.
Last case is $n$ is 2 mod 4. $n=4z+2$. Set $x$ to be $z$. So we will either identify a group of size $2z$, or $2z+2$, i.e. $(n/2)+1$ or $(n/2)-1$.

SIXTH WEIGH: Put ALL THE COINS on one side of the scale. If a real coin weighs $w$ grams, then we will get a number $99w+-$(offset by fakes). We know the offset of the fake coins is even, since there's an even number of fakes! So, we now KNOW the decimal residue of $99w$ mod 2, call it $r$.
$99w$ mod 2 = $r$.
$99(w+-1)$ mod 2 = $r+-99$.
99 is odd, so a 99 fake coins would have a different mod residue to $r$, which we have already identified from our weighing. (Note - i know that 99 fake coins do not necessarily have the same direction of wrong weighting, but the disposition is odd, anyway, making a different residue)

THIS IS A PARTIAL ANSWER. It identifies an even within 9 weighs. I'm posting it here because I believe it covers concepts that could be useful in creating the optimal solution.

THIS IS A PARTIAL ANSWER. It identifies a real coin within 9 weighs. I'm posting it here because I believe it covers concepts that could be useful in creating the optimal solution.