That Was Quick

We look at $2017$ as a number in base $8$ since all numbers less than or equal to $2017_{8}$ do not contain $8$ nor $9$ . Converting back to base $10$ will give us the number of positive integers not greater than $2017$ which do not contain $8$ or $9$ .

$2017_{8}=\boxed{1039_{10}}$

Abiding by the fundamental principles of counting,

Possible single digit numbers: 7 (excluding 8,9)

Possible two digit numbers: $7 \times 8 = 56$

Possible three digit numbers: $7 \times 8 \times 8 = 448$

Possible four digit numbers between 1000 and 1999 inclusive: $1 \times 8 \times 8 \times 8 = 512$

Possible numbers between 2000 and 2017 = 16

Adding them up, $7 + 56 + 448 + 512 + 16 = \boxed{1039}$