Let $n$ be an integer with $1 \le n \le 1000$, and for each integer $i$ with $2 \le i \le 6$, let $r_i$ denote the remainder when $n$ is divided by $i$. We require that $r_2, r_3, r_4, r_5, r_6$ be pairwise distinct. Since $0 \le r_i \le i - 1$, in particular $r_2$ can only be $0$ or $1$.

Case 1: Assume $r_2 = 0$. Then $n$ is even, so $r_4 \in \{0, 2\}$, and $r_2 \neq r_4$ forces $r_4 = 2$. Similarly $r_6 \in \{0, 2, 4\}$ and $r_6 \neq r_2, r_4$ force $r_6 = 4$, which in turn gives $r_3 = 1$. Finally $r_5 \notin \{0, 1, 2, 4\}$ forces $r_5 = 3$. Hence the only possible tuple is
$$(r_2, r_3, r_4, r_5, r_6) = (0, 1, 2, 3, 4),$$
which, by the Chinese remainder theorem, holds precisely when $n \equiv 58 \pmod{60}$.

Case 2: Assume $r_2 = 1$. Then $n$ is odd, so $r_4 \in \{1, 3\}$, and $r_2 \neq r_4$ forces $r_4 = 3$. Similarly $r_6 \in \{1, 3, 5\}$ and $r_6 \neq r_2, r_4$ force $r_6 = 5$, which in turn gives $r_3 = 2$. Finally $r_5 \notin \{1, 2, 3, 5\}$ forces $r_5 \in \{0, 4\}$. Hence the only possible tuples are
$$(r_2, r_3, r_4, r_5, r_6) = (1, 2, 3, 0, 5) \quad \text{and} \quad (1, 2, 3, 4, 5),$$
which, by the Chinese remainder theorem, hold precisely when $n \equiv 35 \pmod{60}$ or $n \equiv 59 \pmod{60}$.

Therefore it suffices to count the integers between $1$ and $1000$ whose remainder modulo $60$ is $35$, $58$, or $59$. There are $17$, $16$, and $16$ such integers, respectively. Hence the total number is $49$.