why does my compare method throw exception -- Comparison method violates its general contract!

I suspect the problem occurs when neither value is sponsored. That will return 1 whichever way you call it, i.e.

x1.compare(x2) == 1

x2.compare(x1) == 1

That's invalid.

I suggest you change this:

object1.getSponsored() && object2.getSponsored()

to

object1.getSponsored() == object2.getSponsored()

in both places. I would probably actually extract this out a method with this signature somewhere:

public static int compare(boolean x, boolean y)

and then call it like this:

public int compare(SRE object1, SRE object2) {
    return BooleanHelper.compare(object1.getSponsored(), object2.getSponsored());
}

That will make the code clearer, IMO.

I assume that you are using JDK 7. Check the following URL:

From http://www.oracle.com/technetwork/java/javase/compatibility-417013.html#source

Area: API: Utilities

Synopsis: Updated sort behavior for Arrays and Collections may throw an IllegalArgumentException

Description: The sorting algorithm used by java.util.Arrays.sort and (indirectly) by java.util.Collections.sort has been replaced. The new sort implementation may throw an IllegalArgumentException if it detects a Comparable that violates the Comparable contract. The previous implementation silently ignored such a situation. If the previous behavior is desired, you can use the new system property, java.util.Arrays.useLegacyMergeSort, to restore previous mergesort behavior.

Nature of Incompatibility: behavioral

RFE: 6804124

For more detailed info, see the bug database reference here.

The contract between equals() and compareTo() is that when equals() returns true, compareTo() should return 0 and when equals() is false compareTo should return -1 or +1.

BTW: I assume your compare() method is not called very often as the debug messages will use up a signficiant amount of CPU and memory.