2016/2017

We have $L = 2^{10} \cdot 3^6 \cdot 7^3 \cdot P,$ where $P$ is a product of other prime powers. Then $L/2016 = 2^5 \cdot 3^4 \cdot 7^2 \cdot P.$ So we are looking for numbers which are divisible by $2^6,$ or $3^5,$ or $7^3.$ No number $\le 2017$ is divisible by two of these simultaneously, so we can count each one separately.

There are $\lfloor 2017 / 2^6 \rfloor = 31$ numbers divisible by $2^6,$ $\lfloor 2017 / 3^5 \rfloor = 8$ numbers divisible by $3^5,$ and $\lfloor 2017 / 7^3 \rfloor = 5$ numbers divisible by $7^3,$ so the answer is $31 + 8 + 5 = \fbox{44}.$