If a number is $ 0,1,$ or $ 2\mod 9$ then say it is the "x" part of a historic set. If it is $ 3,4,$ or $ 5\mod 9$ then say it is the "y" part of a historic set. Finally, if it is $ 6,7,$ or $ 8\mod 9$ then say it is the "z" part of a historic set. Then every nonnegative integer is part of a historic set, because if $ x\equiv 0,1,2\mod 9$ then $ y=2001+x\equiv 3,4,5\mod 9$, and $ z=1776+y\equiv 6,7,8\mod 9$.

tjhance, there is an error in your solution. Your statements imply that numbers like 3,4, and 5 are in y, but for these, there exist no nonnegative x such that x+1776=3,4, or 5. The same applies for z, and for a couple hundred other numbers (all those y under 2001 or those z under 2001+1776). btw...the owner of this username is not the one who is posting this (aka it's not the guy you were punching at ARML).

Drat. I thought it said all integers. I thought that seemed suspiciously easy to not be solved already... EDIT- On second thought, is this even possible? 0 and 1776 must both be the "x" part of a historic set in the partition, because if they were y or z, then the x part would be less than 0. But then 2001+1776 would be the "z" part of the historic set with x=0, but also the "y" part of the historic set with x=1776, a contradiction. Maybe I am misunderstanding again? EDIT EDIT- Oh. You could either have z-y = 1776 and y-x = 2001 OR z-y=2001 and y-x=1776. I can you can just ignore all of my posts.

Set $ a=1776$ and $ b=2001$. We will partition the nonnegative integers into sets by assigning each number as an x, y, or z using the following algorithm: Choose the smallest unassigned integer (call it $ q$) and assign it as an x, and assign $ q+a+b$ as a z. If $ q+a$ is unassigned, then assign it as a $ y$, otherwise assign $ q+b$ as a y. Either way, we will have made a historic set $ \{q,q+a,q+a+b\}$ or $ \{q,q+b,q+a+b\}$. I will show that this algorithm is always possible to complete. All numbers that are currently assigned to $ x$ must be less than $ q$. Therefore, $ q+a+b$ will be unassigned, so we can let it be $ z$. To show that it is possible to assign y to some value, I will show that $ q+b$ must currently be unassigned. It cannot be an x, because it is not less than $ q$. It cannot be a y, because then its associated x value would not be less than $ q$. If it were a z, then $ q+b-(a+b)=q-a$ must be its associated x value. However, when we assigned $ q-a$ to x, we should have assigned $ (q-a)+a=q$ as its y, a contradiction because we are assuming $ q$ is unassigned.