110 000 km
absolute error means what relative error?
I got the 110 000 km value by subtracting my predicted Earth's x coordinate with NASA's Earth x coordinate.
I'm not sure what you're calculating here or what you mean by "NASA's Earth x coordinate". That's a distance from what origin, in what coordinate system, at what time? (As far as I know, the earth moves in orbit around the sun, so its x-coordinate w.r.t. a coordinate system centered at the sun is changing all the time.)
In any case, you calculated an absolute error of 110,000 km by subtracting your calculated value from "NASA's Earth x coordinate". You seem to think this is a bad answer. What's your expectation? To hit it spot on? To be within a meter? One km? What's acceptable to you and why?
You get a relative error by dividing your error difference by "NASA's Earth x coordinate". Think of it as a percentage. What value do you get? If it's 1% or less, congratulate yourself. That would be quite good.
You should know that floating point numbers aren't exact on computers. (You can't represent 0.1 exactly in binary any more than you can represent 1/3 exactly in decimal.) There are going to be errors. Your job as a simulator is to understand the errors and minimize them as best you can.
You could have a stepsize problem. Try reducing your time step size by half and see if you do better. If you do, it says that your results have not converged. Reduce by half again until you achieve acceptable error.
Your equations might be poorly conditioned. Small initial errors will be amplified over time if that's true.
I'd suggest that you non-dimensionalize your equations and calculate the stability limit step size. Your intuition about what a "small enough" step size should be might surprise you.
I'd also read a bit more about the many body problem. It's subtle.
You might also try a numerical integration library instead of your integration scheme. You'll program your equations and give them to an industrial strength integrator. It could give some insight into whether or not it's your implementation or the physics that causes the problem.
Personally, I don't like your implementation. It'd be a better solution if you'd done it with mathematical vectors in mind. The "if" test for the relative positions leaves me cold. Vector mechanics would make the signs come out naturally.
UPDATE:
OK, your relative errors are pretty small.
Of course the absolute error does matter - depending on your requirements. If you're landing a vehicle on a planet you don't want to be off by that much.
So you need to stop making assumptions about what constitutes too small a step size and do what you must to drive the errors to an acceptable level.
Are all the quantities in your calculation 64-bit IEEE floating point numbers? If not, you'll never get there.
A 64 bit floating point number has about 16 digits of accuracy. If you need more than that, you'll have to use an infinite precision object like Java's BigDecimal or - wait for it - rescale your equations to use a length unit other than kilometers. If you scale all your distances by something meaningful for your problem (e.g., the diameter of the earth or the length of the major/minor axis of the earth's orbit) you might do better.