The riddle does not specify that the total size of each pile must be equal, only that each pile has an equal number of tails in each. Make two piles of coins, one containing 90 coins, and the other containing only 10. Then flip all the coins in the pile of 10.
You have 100 coins total, and only 10 are on tails. So the coins can be expressed as follows...
90H 10T
Let's say you pull all 10 tails by some miracle into your tiny little pile of 10. Once you flip them the piles will be as follows
90H and 10H. Both piles have 0 tails.
Now lets say you only pull 1 tails into your pile.
before flip, 81H and 9T / 9H and 1T
After flip 81H and 9T / 1H and 9T, so both piles have 9 coins on tails.
Lets say you grabbed 4 Tails into your pile
before flip 84H 6T / 6H 4T
after flip 84H 6T / 4H 6T, so both piles have 6 coins on tails.
As you can see due to the way the math falls, this would work for any combination of 10 coins.