Solution:

Note $2021 = 43 \cdot 47.$ If $a = \prod p_i^{e_i},$ then $\sigma(a^n) = \prod \sigma(p_i^{e_i n}),$ so it suffices (and is necessary, taking $a$ prime) that $\sigma(p^N) \equiv 1,$ i.e. $p + p^2 + \cdots + p^N \equiv 0 \pmod{43 \cdot 47},$ for every prime $p$ and every multiple $N$ of $n.$

Fix $q \in \{43, 47\}.$ If $q \mid p$ the sum is $0.$ If $p \equiv 1 \pmod q$ the sum is $\equiv N,$ so choosing such a prime (Dirichlet) forces $q \mid n.$ Otherwise the sum is $p \cdot \frac{p^N - 1}{p - 1}$ with $p - 1$ invertible, so we need $p^N \equiv 1 \pmod q;$ choosing $p$ to be a primitive root mod $q$ forces $q - 1 \mid n.$ Conversely, if $q(q-1) \mid n$ then for every multiple $N$ of $n$ and every prime $p,$ the sum vanishes mod $q$ in all three cases. Hence the least $n$ is $$n = \operatorname{lcm}(43 \cdot 42,\ 47 \cdot 46) = 2 \cdot 3 \cdot 7 \cdot 23 \cdot 43 \cdot 47.$$

The sum of the prime factors is $2 + 3 + 7 + 23 + 43 + 47 = 125.$