Solution:
Let $f(n)$ be the cyclostyled of $n$.

a.
Let us call $s(m)$ the sum of the digits of $m$. By the divisibility criterion for 9 we have that $f(n)$ is a multiple of 9 if and only if $s(f(n))$ is a multiple of 9; on the other hand $s(f(n)) = 2012 \cdot s(n)$, which is a multiple of 9 if and only if $s(n)$ is (because the greatest common divisor between 2012 and 9 is equal to 1). Using again the divisibility criterion for 9 we observe that $s(n)$ is a multiple of 9 if and only if $n$ is.

In conclusion the $n$ sought are all and only the multiples of 9.

b.
Let us call $r(m)$ the alternating sum of the digits of $m$ (done in such a way that the units digit is taken with a positive sign). By the divisibility criterion for 11 we have that $f(n)$ is a multiple of 11 if and only if $r(f(n))$ is a multiple of 11.

We distinguish the following two cases.

- $n$ has an even number of digits.

Then $r(f(n)) = 2012 \cdot r(n)$, indeed each digit of $n$ is summed 2012 times with the same sign. Since $\operatorname{GCD}(2012,11) = 1$ we observe that $2012 \cdot r(n)$ is a multiple of 11 if and only if $r(n)$ is. Using again the divisibility criterion for 11 we have that $r(n)$ is a multiple of 11 if and only if $n$ is.

Therefore the $n$ with an even number of digits that we want are all and only the multiples of 11.

- $n$ has an odd number of digits.

In the alternating sum of the digits of $f(n)$ we have that each digit of $n$ is summed $\frac{2012}{2}$ times with the sign $+$ and just as many times with the sign $-$; therefore $r(f(n)) = 0$, that is, $f(n)$ is always a multiple of 11.

In conclusion the $n$ sought are the positive integers with an odd number of digits and the multiples of 11 with an even number of digits.