I'm unsure as to how $I=0$ mod $3$ combined with $a+b+c=$ mod $3$ is the condition for the monochromatic state. I've thought about this problem backwards and realize that if we wish to have $c$ be the sole remaining color, then the state preceding this state would be for there to be $1$ of $b$ and $1$ of $a$. So that sort of explains why we'd need $I(a,b,c)=a-b$ mod$3=0$ mod $3$ but why do we need $a+b+c=$ mod $3$? I'd appreciate a big picture clarification on what's going on in this proof.
Thanks! $\phantom{}$
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- Solution
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The given solution is incorrect, so it's no wonder you don't understand it. Since $a-b\pmod3,a-c\pmod3,\text{ and }b-c\pmod3$ are invariant, at least one of these must be $0$ in order to be able to reach a state where two of $a,b,c$ are $0$. There is no requirement that $a+b+c\equiv0\pmod3,$ however. The state $(2,0,0)$ has all chips the same color, after all.
This shows that the condition is necessary, but not that it's sufficient. One has to give an algorithm to show that if $a\equiv b\pmod3,$ fior example, then it is possible to reduce to a state where $a=b=0$. This isn't difficult, and I'll leave it to you.
As for the last question in the problem, "What states can be reached from these numbers," I'm not certain. I believe the answer is all states where $a+b+c=45$ and the invariants are preserved, but I've never actually proved this. Again, it's obvious that these are the only possible states, but one has to give an algorithm to get from one state to another. I'd imagine it's just a simple elaboration of the algorithm to get to a monochromatic state.
EDIT
Here's the way to reduce $(9,3,1)$ to $(0,0,13)$ that the OP mentioned in a comment: $$(9,3,1)\to(8,5,0)\to(7,4,2)\to(6,6,1)$$ and from there you just reduce $a$ and $b$.