One of your other entries has been linked to on Reddit. I am just browsing around. You write well and education is a topic about which I like to read so Kudos and thank you!
The problem and solution here are very interesting. May I suggest another solution?
The digits 1, 2, 3, and 4 arranged in any order will always result in a 4-digit number that has a remainder of 1 when divided by 3. The sum of 4 copies of these numbers will have a remainder of 1 when dividing by 3. Unfortunately 9000 is divisible by 3 or has a remainder of zero so the sum of the 4 copies cannot be 9000.
I wonder how close one can get to 9000 though. Food For Thought.