Solution:

Let $A$ be the set of numbers between $1$ and $2016$ divisible by $4$.
Let $B$ be the set of numbers divisible by $6$.
Let $C$ be the set of numbers divisible by $10$.

We are to find $|A \setminus (B \cup C)| + |B \setminus (A \cup C)| + |C \setminus (A \cup B)|$.

First, compute the sizes of the sets:

$|A| = \left\lfloor \dfrac{2016}{4} \right\rfloor = 504$

$|B| = \left\lfloor \dfrac{2016}{6} \right\rfloor = 336$

$|C| = \left\lfloor \dfrac{2016}{10} \right\rfloor = 201$

Now, intersections:

$|A \cap B|$ is numbers divisible by $\operatorname{lcm}(4,6) = 12$:
$|A \cap B| = \left\lfloor \dfrac{2016}{12} \right\rfloor = 168$

$|A \cap C|$ is numbers divisible by $\operatorname{lcm}(4,10) = 20$:
$|A \cap C| = \left\lfloor \dfrac{2016}{20} \right\rfloor = 100$

$|B \cap C|$ is numbers divisible by $\operatorname{lcm}(6,10) = 30$:
$|B \cap C| = \left\lfloor \dfrac{2016}{30} \right\rfloor = 67$

$|A \cap B \cap C|$ is numbers divisible by $\operatorname{lcm}(4,6,10) = 60$:
$|A \cap B \cap C| = \left\lfloor \dfrac{2016}{60} \right\rfloor = 33$

Now, for exactly one:

Numbers divisible by $4$ only:
$|A| - |A \cap B| - |A \cap C| + |A \cap B \cap C| = 504 - 168 - 100 + 33 = 269$

Numbers divisible by $6$ only:
$|B| - |A \cap B| - |B \cap C| + |A \cap B \cap C| = 336 - 168 - 67 + 33 = 134$

Numbers divisible by $10$ only:
$|C| - |A \cap C| - |B \cap C| + |A \cap B \cap C| = 201 - 100 - 67 + 33 = 67$

Total: $269 + 134 + 67 = 470$

Answer: $470$