Besides bounding the number of steps, this is a nice argument to reduce the arbitrary even $n$ case to $n=6$.

I took the liberty of cleaning up some formatting and correcting several small inaccuracies. I did this to promote this answer as a correct solution of the real-entry case, as it seems to be overlooked (as of writing, I am the only up-voter).

$N_0$ is called the FC-center of $G$. groupprops.subwiki.org/wiki/FC-center

@BillyJoe Suppose the initial list is $(x_1, \dots, x_n)$ where $x_i > 0$ for all $i$, and suppose some number of moves changes the list to $(N, \dots, N)$. Any move that creates an $N$ has the form $(x, y) \to (x+y, x+y)$ where $x, y> 0$ and $x+y = N$, so it creates precisely two $N$'s. Moreover no move ever destroys an $N$ (because that would create a bigger element). Therefore the initial list $(x_1, \dots, x_n)$ must contain an odd number of $N$'s. In particular for $(2, \dots, 2, 1)$ the only possibility is $N=1$, but that's impossible because it's less than $2$.

Hmm, it's no longer clear to me how to generalize to all even $n$, but any multiple of $6$ is fine, because if $n = 6m$ then you can inductively reduce to the form $P(x, y) = x^{4m} y^{2m}$, and then you have reductions $P(x, y) \to P(x, 2y) \to P(x/2, y)$ when $x$ is even, $P(x, y) \to P(2x, y) \to P(x, y/2)$ when $y$ is even, and also $P(x, y) \to P(x+y, x) \to P((x+y)/2, x)$ when $x$ and $y$ are both odd. What if $n = 10$?

An argument that odd is always impossible is sketched at math.stackexchange.com/questions/4746506/…. (You argue that if everything is positive and the ending is $(n,\dots,n)$ then there must be an odd number of $n$'s on the initial list, so for example $(2,\dots,2,1)$ can't be solved.)

The same argument does not work for real numbers though, so that question remains open. But observe that we may straight away reduce to $(x, x, x, x, y, y)$, and thereafter all entries remain in $\mathbf Z x + \mathbf Z y$, so it is essentially equivalent to ask about the single case $(e_1, e_1, e_1, e_1, e_2, e_2)$ where $e_1$ and $e_2$ are the basis vectors of $\mathbf Z^2$. The argument above breaks down because elements of $\mathbf Z^2$ really have four kinds of parity.

The same argument works for any even number in place of $6$. What about odd numbers?

$q_i$ is an element of $\mathbb R/\mathbb Z$, not $C_i$. In $\mathbb R/\mathbb Z$ there are always exactly two solutions to $2x=y$. If $y$ has finite odd order $k$ then one of the solutions has order $k$ and the other has order $2k$.

@AndreaAntinucci For $q$ to be well-defined the expression on the right must depend only on $n_i$ modulo $r_i$. If you replace $n_i$ with $n_i + r_i$ the difference is $2n_i r_i q_i + r_i^2 q_i + \sum_{j \ne i} \chi(g_i, g_j) r_i n_j = r_i^2 q_i$, and this must be zero.

Incidentally I was expecting to see the requirement $q(na) = n^2 q(a)$ for all $n \in \mathbb Z$, but I see this actually follows from $q(-a) = q(a)$.

Question 2 is trivial: it is the set of all functions $q :A \to \mathbb R / \mathbb Z$ such that $q(-a) = q(a)$ and $q(a+b) = q(a) + q(b)$ identically, which is obviously $\mathrm{Hom}(A, \frac12 \mathbb Z / \mathbb Z)$.

E.g., if $k = 1$, then $C_{k,n} = \{1,\dots,n\}$. What kind of "algebraic inner structure" can you expect?