Let us prove the following
Statement: $n = k^2 - k + 1$ for some positive integer $k$.
Proof: Let $d$ be the largest of the greatest common divisors found. Then the equality $\gcd(a_i, a_{i+\ell}) = d$ is equivalent to the fact that each of the numbers $a_i$ and $a_{i+\ell}$ is a multiple of $d$. Let $(i_1, i_2, \dots, i_k)$ be all indexes of numbers in the given sequence that are multiples of $d$, written in ascending order. Note that each pair $i_j < i_s$ of indices yields $d$ twice: for $\ell = s - j$ and for $\ell = j + n - s$. Therefore $d$ will occur as the greatest common divisor exactly $2\left(\frac{k(k-1)}{2}\right)$ times and by assumption this number is $n-1$. Thus $n = 1 + 2\left(\frac{k(k-1)}{2}\right) = k^2 - k + 1$. The assertion is proved.

b) The equation $k^2 - k + 1 = 2021$ is equivalent to $k(k-1) = 2020$. The left side of this equation increases for $k > 0.5$ while $90 \cdot 89 = 8010$ and $91 \cdot 90 = 8190$, therefore this equation doesn't have a positive integer solutions. According to the proven statement, the answer at this item is: "no".

a) Note that $21 = 5^2 - 5 + 1$, i.e. $n = 21$ satisfies the statement for $k = 5$. It is easy to construct the sequence $(i_1, i_2, i_3, i_4, i_5)$ from the proof of this statement, for example $(1, 11, 13, 18, 19)$. Let us set $a_1 = a_2 = a_3 = \dots = a_{21} = 1$ then choose $20$ different primes $p_1 > p_2 > \dots > p_{20}$ and for each $i$ from $1$ to $20$ multiply by $p_i$ the numbers $a_i, a_{i+10}, a_{i+12}, a_{i+17}, a_{i+18}$. It is easy to see that for each $\ell$ from $1$ to $20$ the set of greatest common divisors from the problem statement coincides with the set $\{p_1, p_2, \dots, p_{20}\}$.