The remainder mod $2$ is forced by the remainders mod $4$ and mod $6,$ and the remainder mod $3$ by the one mod $6$ — few patterns survive.

Everything depends only on $n \bmod 60;$ exactly three residues work, so count each class below $1000.$

Write $r_k$ for the remainder of $n$ modulo $k,$ and note $r_2 \equiv r_4 \pmod 2,$ $r_2 \equiv r_6 \pmod 2,$ and $r_3 \equiv r_6 \pmod 3.$ If $r_2 = 0:$ then $r_4$ is even and different from $r_2,$ so $r_4 = 2;$ then $r_6$ is even and avoids $\{0, 2\},$ so $r_6 = 4,$ which gives $r_3 = 1;$ finally $r_5$ avoids $\{0, 1, 2, 4\},$ so $r_5 = 3.$ These say $n \equiv -2$ modulo each of $2, 3, 4, 5, 6,$ i.e. $n \equiv 58 \pmod{60}.$

If $r_2 = 1:$ similarly $r_4 = 3,$ then $r_6 = 5,$ giving $r_3 = 2,$ and $r_5$ avoids $\{1, 2, 3, 5\},$ so $r_5 \in \{0, 4\}.$ The choice $r_5 = 4$ gives $n \equiv -1 \pmod{60},$ i.e. $n \equiv 59;$ the choice $r_5 = 0$ gives $n \equiv 11 \pmod{12}$ with $5$ dividing $n,$ i.e. $n \equiv 35 \pmod{60}.$

Below $1000$ there are $17$ integers congruent to $35,$ $16$ congruent to $58,$ and $16$ congruent to $59$ modulo $60,$ for a total of $17 + 16 + 16 = 49.$