Imagine there is a cinema hall and there are $n$ seats and we want to arrange $n$ people with some special conditions on our seats.
Each people have number from $1$ to $n$ and clearly our seats is numbered form $1$ to $n$ either and now consider these conditions.
$\color{blue}i)\color{blue}{\text{Each people can sit on his number seat and even in his multiple seats}}$, for example, person with number $1$ can sit in all of our seats and person with number $3$ can sit on seats with number $3, 6, 9\dots$
Now We want to $ \color{blue}{\text{arrange these people in the way that minimize seated people}}$ in this cinema.
For example, our cinema have $6$ seat's and we have $6$ people (note that the number of people is the same with seats)so in order to minimize seated people we have to arrange people in this way. Person number $1$ has to sit in seat number $5$ and person number $2$ has to sit in seat number $4$ person number $3$ has to seat in seat number $6$ in this way persons numbers $4, 5, 6$ can't in the cinema because their seats are full.
So how can I find the minimum number of seated people in the cinema and minimum number of seats required if a condition is given as if my above example?
I can keep out $37\%$ of people this way:
1. For $N/16<m\leq N/8$, put $m$ in seat $4m$ and $2m$ in seat $8m$.
2. For $N/4<m\leq N/2, 4\nmid m$, put $m$ in seat $2m$.
3. For odd $N/6<m\leq N/4$, put $m$ in seat $3m$.
4. For odd $3N/20<m\leq N/6$, put $m$ in seat $5m$.
5. For odd $N/8<m\leq3N/20,3\nmid m$, put $m$ in seat $5m$.
6. For odd $N/8<m\leq N/7, 3\mid m$, put $m$ in seat $7m$.
This unseats $2/16+3/16+1/24+1/120+1/120+1/336=157/420$ of the people.
You still have the first $N/16$ people to seat as well.
You can't do better than $1/2$ because everyone shut out has someone sitting in their chair.