Consider the following cases:

(1) If there exists $a$ such that $f(a) = a$, then $f(a) + f^{-1}(a) = 2a \neq 2013$. So this is impossible.

(2) If there exist $a \neq b$ such that $f(a) = b$ and $f(b) = a$, then $f(a) + f^{-1}(a) = 2b \neq 2013$. So this is impossible.

(3) If there exist three distinct numbers $a, b, c$ such that $f(a) = b, f(b) = c$ and $f(c) = a$, then $f(a) + f^{-1}(a) = b + c$ and $f(b) + f^{-1}(b) = c + a$. Since $b + c = c + a = 2013$, we get $b = c$, which contradicts the assumption.

(4) If there exist $k$ (with $k \ge 5$) distinct numbers $a_1, a_2, \dots, a_k$ such that $f(a_1) = a_2, f(a_2) = a_3, \dots, f(a_{k-1}) = a_k$ and $f(a_k) = a_1$, then $f(a_2) + f^{-1}(a_2) = a_3 + a_1 = 2013 = f(a_4) + f^{-1}(a_4) = a_5 + a_3$, which gives $a_1 = a_5$, contradicting the assumption.

(5) If there exist 4 distinct numbers $a, b, c, d$ such that $f(a) = b, f(b) = c, f(c) = d$ and $f(d) = a$, then $f(b) + f^{-1}(b) = a + c = 2013 = f(d) + f^{-1}(d)$, i.e., $c = 2013 - a$; also $f(a) + f^{-1}(a) = b + d = 2013 = f(c) + f^{-1}(c)$, i.e., $d = 2013 - b$. Therefore $(a, b, c, d) = (a, b, 2013 - a, 2013 - b)$ forms a cyclic group.

Since only (5) can hold, it suffices to determine 503 cyclic groups $(a_i, b_i, 2013-a_i, 2013-b_i)$ to determine such a function $f$. Since $a_i$ and $2013-a_i$ respectively have one less than 1007 and one greater than 1007; the same holds for $b_i$ and $2013-b_i$; so we may assume $1 \le a_i, b_i \le 1006$ in order to determine $(a_i, b_i, 2013-a_i, 2013-b_i)$ (at this point it is no longer a cyclic group, but an ordered group). Choosing $a_1,b_1,a_2,b_2,\cdots,a_{503},b_{503}$ in order from $\{1,2,\cdots,1006\}$, there are $1006!$ possibilities; however, these 503 groups have no order among themselves, so the number of $f$ is:
$$\frac{1006!}{503!}$$