The values of $j$ in question are those not divisible by either $42$ or $46$.

We first check that for $p$ prime,
$$\sum_{n=1}^{p-1} n^j \equiv 0 \pmod{p} \Leftrightarrow j \not\equiv 0 \pmod{p-1}.$$
If $j \equiv 0 \pmod{p-1}$, then $n^j \equiv 1 \pmod{p}$ for each $n$, so $\sum_{n=1}^{p-1} n^j \equiv p-1 \pmod{p}$. If $j \not\equiv 0 \pmod{p-1}$, we can pick a primitive root $m$ modulo $p$,
observe that $m^j \not\equiv 1 \pmod{p}$, and then note that
$$\sum_{n=1}^{p-1} n^j \equiv \sum_{n=1}^{p-1} (mn)^j = m^j \sum_{n=1}^{p-1} n^j \pmod{p},$$
which is only possible if $\sum_{n=1}^{p-1} n^j \equiv 0 \pmod{p}$.

We now note that the prime factorization of 2021 is $43 \times 47$,
so it suffices to determine when $S(j)$ is divisible by each of 43 and 47.
We have
\begin{align*}
S(j) &\equiv 46 \sum_{n=1}^{42} n^j \pmod{43} \\
S(j) &\equiv 42 \sum_{n=1}^{46} n^j \pmod{47}.
\end{align*}
Since 46 and 42 are coprime to 43 and 47, respectively,
we have
\begin{gather*}
S(j) \equiv 0 \pmod{43} \Leftrightarrow j \not\equiv 0 \pmod{42} \\
S(j) \equiv 0 \pmod{47} \Leftrightarrow j \not\equiv 0 \pmod{46}.
\end{gather*}
This yields the claimed result.