It is possible.
Guess the color you would have, assuming that there are an odd number of red hats. If the number of red hats is odd, everyone will be correct. If it is even, everyone will be wrong.
All winning strategies:
only two winning strategies: the one given above, or assuming an even number of red hats.
Choose a strategy S where always either everyone is right or everyone is wrong.
Claim: If S has everyone right when all hats are blue, then everyone is correct when there are an even number of red hats and everyone is wrong when there are an odd number of red hats.
Proof by induction on number of red hats. We are told it is true for 0 red hats.
Suppose it is true for $k-1$ red hats. Consider a case where exactly $k$ hats are red, with $k$ even. Choose a person R with a red hat. R sees exactly the same thing as when everyone else has the same, but R has a blue hat, so R must guess the same. In that second case, there are $k-1$ red hats (an odd number), so R guessed wrong, saying "red". In the case when R is wearing a red hat, R guessed the same, so R guessed correctly. Since S has everyone right or everyone wrong, everybody is right with $k$ red hats.
Similarly, with $k$ hats red, $k$ odd, then the case with R wearing blue but everyone else the same had $k-1$ red hats ($k-1$ even), so R correctly guessed blue. With R wearing a red hat, R must have guessed the same (blue), which is now wrong, so everyone is wrong with $k$ red hats.
Proof when everyone is wrong with 0 red hats goes similarly.
