jmc

We first show that we can choose at most 5 numbers from $\{1,2,\dots ,11\}$ such that no two numbers have a difference of $4$ or $7$. We take the smallest number to be $1$, which rules out $5,8$. Now we can take at most one from each of the pairs: $[2,9]$, $[3,7]$, $[4,11]$, $[6,10]$. Now, $1989=180\cdot 11+9$. Because this isn't an exact multiple of $11$, we need to consider some numbers separately. Notice that $1969=180\cdot 11-11=179\cdot 11$. Therefore we can put the last $1969$ numbers into groups of 11. Now let's examine $\{1,2,\dots ,20\}$. If we pick $1,3,4,6,9$ from the first $11$ numbers, then we're allowed to pick $11+1$, $11+3$, $11+4$, $11+6$, $11+9$. This means we get 10 members from the 20 numbers. Our answer is thus $179\cdot 5+10=\boxed{905}$.