Let $N = 2011$.

Let $A$ be the set of positive integers less than $2011$ divisible by $2$ or $7$.
Let $B$ be the set of positive integers less than $2011$ divisible by $5$.
We are to find $|A \setminus B|$.

First, count the number of positive integers less than $2011$ divisible by $2$ or $7$.

Let $S_2$ be the set divisible by $2$.
Let $S_7$ be the set divisible by $7$.
Let $S_{14}$ be the set divisible by $14$.

Number divisible by $2$: $\left\lfloor \dfrac{2010}{2} \right\rfloor = 1005$
Number divisible by $7$: $\left\lfloor \dfrac{2010}{7} \right\rfloor = 287$
Number divisible by $14$: $\left\lfloor \dfrac{2010}{14} \right\rfloor = 143$

By inclusion-exclusion:
Number divisible by $2$ or $7$ is $1005 + 287 - 143 = 1149$.

Now, subtract those divisible by $5$.

Let $S_{2,5}$: divisible by $2$ and $5$ ($10$)
$\left\lfloor \dfrac{2010}{10} \right\rfloor = 201$

Let $S_{7,5}$: divisible by $7$ and $5$ ($35$)
$\left\lfloor \dfrac{2010}{35} \right\rfloor = 57$

Let $S_{14,5}$: divisible by $14$ and $5$ ($70$)
$\left\lfloor \dfrac{2010}{70} \right\rfloor = 28$

By inclusion-exclusion, the number of integers less than $2011$ divisible by $2$ or $7$ and also by $5$ is:
$201 + 57 - 28 = 230$

Therefore, the answer is $1149 - 230 = 919$.

Answer: $919$