This is a companion to my earlier question, Sequences with integral means. This new question is, frankly, not as interesting, but it feels necessary to complete the thought.

Let $V(n)$ be the sequence whose first element is $n$, and from then onward, the next element is the smallest natural number ${\ge}1$ that ensures that the variance of all the numbers in the sequence is an integer. If there is no such number to extend $V(n)$, then it has finite length.

For example, for $n=20$, $V(20) = 20,2,2,4$.
When the last element $4$ is added, the mean is $28/4=7$, and the
variance sum is
$$13^2 + (-5)^2 + (-5)^2 + (-3)^2 = 169+25+25+9 = 228$$
which is divisible by $n=4$ yielding a variance of $57$.
(*NB*: An earlier version of this question divided by $n{-}1$ rather than by $n$; see the comments.)
None of the smaller alternatives $1,2,3$ lead to integral variance.
The $V(20)$ sequence cannot be extended beyond that terminating $4$,
for the attempt to extend leads to a quadratic Diophantine equation with no solution.

Q1. Is it the case that the only infinite sequences are $V(1)=1,1,1,\ldots$ and $V(2)=2,2,2,\ldots$ ?

Q2. Is there nevertheless no upperbound on the length of the longest finite sequence?

Here are some "long" sequences encountered within $n \le 1000$, of lengths $6$, $6$, $8$, and $9$ respectively: $V(61)=61,1,1,1,1,1$; $V(62)=62,2,2,2,2,2$; $V(422)=422,2,2,2,2,2,2,6$; $V(842)=842,2,2,2,2,2,2,2,8$.