Let $r_2, r_3,\ldots, r_{1000}$ denote the remainders when a positive odd integer is divided by $2,3,\ldots,1000$, respectively. It is known that the remainders are pairwise distinct and one of them is $0$. Find all values of $k$ for which it is possible that $r_k = 0$.
1. Let \( r_2, r_3, \ldots, r_{1000} \) denote the remainders when a positive odd integer \( N \) is divided by \( 2, 3, \ldots, 1000 \), respectively. It is given that these remainders are pairwise distinct and one of them is \( 0 \).

2. Assume \( N \in \mathbf{S} \), where \( \mathbf{S} \) is the set of integers satisfying the given conditions. Let \( p \) be the unique integer for which \( r_p = 0 \).

3. If \( p \) is composite, there exists a prime \( q < p \) which divides \( p \), implying \( r_q = 0 \). This is a contradiction since the remainders are pairwise distinct. Therefore, \( p \) must be a prime number.

4. Since \( N \) is odd, \( r_2 = 1 \). This implies \( p \neq 2 \).

5. The fact that \( 0 \leq r_k \leq k-1 \) implies \( r_3 = 2 \), \( r_4 = 3 \), \( r_5 = 4 \), and so on, up to \( r_{p-1} = p-2 \). This gives us the sequence:
   \[
   r_k = k-1 \quad \text{for} \quad 2 \leq k \leq p-1.
   \]

6. Assume \( p < 500 \). Then \( 2p < 1000 \). Since \( r_p = 0 \), \( p \) divides \( r_{2p} \), implying \( r_{2p} = p \) because \( r_{2p} < 2p \) and \( r_{2p} \neq r_p = 0 \).

7. This combined with the sequence \( r_k = k-1 \) for \( 2 \leq k \leq p-1 \) and the fact that \( r_{p+1} \in \{0, 1, 2, \ldots, p\} \) implies \( r_{p+1} = p-1 \). However, \( r_{p+1} \) must be odd since \( p \) is an odd prime, leading to a contradiction. Therefore, \( p > 500 \).

8. Next, assume \( 500 < p < 1000 \). For positive integers \( k \leq p \), define the odd positive integer \( N_k = \frac{1000!}{p}k - 1 \).

9. Let \( m \) be an integer such that \( 2 \leq m \leq 1000 \) and \( m \neq p \). Clearly, \( m \) divides \( \frac{1000!}{p} \), hence:
   \[
   N_k \equiv -1 \equiv m-1 \pmod{m},
   \]
   i.e., the remainder is \( m-1 \) when \( N_k \) is divided by \( m \).

10. Assume \( a \) and \( b \) are integers such that \( 1 \leq a < b \leq p \) and \( N_a \equiv N_b \pmod{p} \). This implies:
    \[
    \frac{1000!}{p}a - 1 \equiv \frac{1000!}{p}b - 1 \pmod{p} \implies \frac{1000!}{p}(b-a) \equiv 0 \pmod{p}.
    \]
    Since \( 1 \leq a < b \leq p \), \( 0 < b - a < p \), hence \( p \nmid b-a \). Also, \( p \nmid \frac{1000!}{p} \) because \( p > 500 \) implies \( 1000 < 2p < p^2 \), meaning \( p^2 \nmid 1000! \). Consequently, \( N_a \not\equiv N_b \pmod{p} \).

11. Therefore, \( N_1, N_2, \ldots, N_p \) are congruent to \( 0, 1, 2, \ldots, p-1 \) modulo \( p \) in some order. Thus, there is one unique integer \( c \in \{1, 2, \ldots, p\} \) such that \( N_c \equiv 0 \pmod{p} \).

12. Summarizing, \( N_c \) has 999 distinct remainders when divided by \( 2, 3, \ldots, 1000 \) and one of them is zero. Hence, \( N_c \in \mathbf{S} \), leading to the conclusion:

The values of \( k \) for which it is possible that \( r_k = 0 \) are all primes between 500 and 1000.

The final answer is \( \boxed{\text{all primes between 500 and 1000}} \).