Problem 1417

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 $\boxed{\text{all primes between 500 and 1000}}$.