For every integer $x$, denote by $G(x)$ the number into which Gandalf turns the number $x$.

a. Suppose that both $1001$ and $1003$ are reflecting. Then, for every integer $x$, we have $G(x+2) = G(1003 - (x+2)) = G(1001 - x) = G(x)$. Hence Gandalf turns all even numbers into one and the same integer $c$ and all odd numbers into one and the same integer $c'$. But $c = G(500) = G(501) = c'$, implying that all integers are turned into one and the same integer. This contradicts the assumption that Gandalf turns each integer into some other integer.

b. Suppose that numbers $1000$, $1003$ and $1008$ are all reflecting. Then, for every integer $x$,
$$G(x+3) = G(1003 - (x+3)) = G(1000 - x) = G(x),$$
$$G(x+5) = G(1008 - (x+5)) = G(1003 - x) = G(x).$$
So $G(x+1) = G(x+4) = G(x+7) = G(x+10) = G(x+5) = G(x)$ for every integer $x$. Consequently, Gandalf again turns all integers into equal integers, contradicting the condition of the problem.

c. Suppose Gandalf turns all even numbers into $1$ and all odd numbers into $2$. Then no integer is left unchanged. For every even number $a$, including $1002$, $1004$ and $1006$, the numbers $x$ and $a-x$ are either both even or both odd, whence they are turned into equal numbers.