Just for kicks, I went out and took some measurments off my tractor. The distance from the center of the front axle to the bucket edge is 67". The distance from front axle to 3PH pins is 91". 67/91=.74.
What that means is to maintain the same F/R weight % you would need 74% of whatever your loader can lift hanging off the 3PH pins. Assuming your tractors measurments equal about the same ratio. Your loader can lift~ 545lbs at the edge. 545*.74=~400lbs of ballast needed to maintain the same weight distribution.
But 500-600 would be a little better IMO because even with no load in the bucket, if you rais it all the way, it can become a little unstable. 2 reasons is that it raises the center of gravity, and also because it moves the load farther from the front axle which is what the tractor tries to pivot around. Thus giving it more leverage on the ballast.
I also wanted to point out that my wheel base is 64". The loader pins are 67" in front of the front axle. Almos a 1:1 ratio. That means that for every pound of fluid in the tire will only ballast 1 pound in the bucket. I have loaded tires @ 200lbs each on a loader that will lift 853lbs @ the edge. Loaded rears will only account for about half the load in the bucket.
Just some food for though for those who care to ponder.