Answer: 905

Small Hint:

Within any eleven consecutive integers, the forbidden-difference graph is an $11$-cycle

Big Hint:

For a matching construction, look for five allowable residue classes modulo $11$

Solution:

On any eleven consecutive integers, join two numbers when their difference is $4$ or $7.$ Because $7=11-4,$ this graph is an $11$-cycle, whose largest independent set has size $5.$ Similarly, any ten consecutive integers induce a path on ten vertices and contribute at most $5.$ Since $1989=179\cdot11+2\cdot10,$ this gives $|S|\leq179(5)+2(5)=905.$

This bound is attained by taking every integer whose residue modulo $11$ is $1,$ $3,$ $4,$ $6,$ or $9.$ No two selected residues differ by $4$ or $7$ modulo $11,$ and each of these five residues occurs $181$ times from $1$ through $1989.$ Thus the maximum is $5(181)=905.$