, 2013

Solution:
First, we can let $h=\operatorname{gcd}(a, b)$ so that $(a, b)=(h A, h B)$ where $\operatorname{gcd}(A, B)=1$. Making these substitutions yields
$$\frac{k}{2013}(h A+h B)=h A B,$$
so
$$k=\frac{2013 A B}{A+B}.$$
Because $A$ and $B$ are relatively prime, $A+B$ shares no common factors with either $A$ or $B$, so in order to have $k$ be an integer, $A+B$ must divide $2013$, and since $A$ and $B$ are positive, $A+B>1$.

We first show that for different possible values of $A+B$, the values of $k$ generated are distinct. In particular, we need to show that
$$\frac{2013 A B}{A+B} \neq \frac{2013 A' B'}{A'+B'}$$
whenever $A+B \neq A'+B'$. Assume that such an equality exists, and cross-multiplying yields
$$A B (A'+B') = A' B' (A+B).$$
Since $A B$ is relatively prime to $A+B$, we must have $A+B$ divide $A'+B'$. With a similar argument, we can show that $A'+B'$ must divide $A+B$, so $A+B=A'+B'$.

Now, we need to show that for the same denominator $A+B$, the values of $k$ generated are also distinct for some relatively prime non-ordered pair $(A, B)$. Let $n=A+B=C+D$. Assume that
$$\frac{2013 A B}{n}=\frac{2013 C D}{n},$$
or equivalently, $A(n-A)=C(n-C)$. After some rearrangement, we have $(C+A)(C-A)=n(C-A)$. This implies that either $C=A$ or $C=n-A=B$. But in either case, $(C, D)$ is some permutation of $(A, B)$.

Our answer can therefore be obtained by summing up the totients of the factors of $2013$ (excluding $1$) and dividing by $2$ since $(A, B)$ and $(B, A)$ correspond to the same $k$ value, so our answer is
$$\frac{2013-1}{2}=1006.$$