Let $A$ be the set of all integers $n$ such that $1 \le n \le 2021$ and $\text{gcd}(n,2021)=1$. For every nonnegative integer $j$, let
\[
S(j)=\sum_{n \in A}n^j.
\]
Determine all values of $j$ such that $S(j)$ is a multiple of $2021$.
1. **Identify the problem and given conditions:**
   We are given a set \( A \) of integers \( n \) such that \( 1 \leq n \leq 2021 \) and \(\text{gcd}(n, 2021) = 1\). We need to determine all values of \( j \) for which \( S(j) = \sum_{n \in A} n^j \) is a multiple of 2021.

2. **Factorize 2021:**
   \[
   2021 = 43 \times 47
   \]
   Since 43 and 47 are prime, we can use the Chinese Remainder Theorem (CRT) to simplify the problem.

3. **Apply the Chinese Remainder Theorem:**
   By CRT, \( S(j) \) is a multiple of 2021 if and only if:
   \[
   S(j) \equiv 0 \pmod{43} \quad \text{and} \quad S(j) \equiv 0 \pmod{47}
   \]

4. **Consider the set \( A \) modulo 43 and 47:**
   - For \( \text{mod} \ 43 \):
     The integers \( n \) in \( A \) that are coprime to 43 are \( 1, 2, \ldots, 42 \). Thus, we need:
     \[
     \sum_{i=1}^{42} i^j \equiv 0 \pmod{43}
     \]
   - For \( \text{mod} \ 47 \):
     The integers \( n \) in \( A \) that are coprime to 47 are \( 1, 2, \ldots, 46 \). Thus, we need:
     \[
     \sum_{i=1}^{46} i^j \equiv 0 \pmod{47}
     \]

5. **Use properties of sums of powers modulo a prime:**
   It is known that for a prime \( p \), the sum of the \( j \)-th powers of the first \( p-1 \) integers is zero modulo \( p \) if and only if \( p-1 \nmid j \). This is a result from properties of sums of powers and Fermat's Little Theorem.

6. **Apply this property to our specific primes:**
   - For \( p = 43 \):
     \[
     \sum_{i=1}^{42} i^j \equiv 0 \pmod{43} \quad \text{if and only if} \quad 42 \nmid j
     \]
   - For \( p = 47 \):
     \[
     \sum_{i=1}^{46} i^j \equiv 0 \pmod{47} \quad \text{if and only if} \quad 46 \nmid j
     \]

7. **Combine the conditions:**
   Therefore, \( S(j) \) is a multiple of 2021 if and only if:
   \[
   42 \nmid j \quad \text{and} \quad 46 \nmid j
   \]

The final answer is \( \boxed{42 \nmid j, \ 46 \nmid j} \)