2013 is factorized as $2013 = 3 \times 11 \times 61$. Let $p_1 = 3$, $p_2 = 11$, $p_3 = 61$. We denote by $a_i$ the residue of $a$ modulo $p_i$, by $b_i$ the residue of $b$ modulo $p_i$ ($i = 1, 2, 3$), $a, b \in \{1, 2, \dots, 2013\}$. By the Chinese Remainder Theorem, we have a bijection of $(a, b)$ with $(a_1, a_2, a_3, b_1, b_2, b_3)$.

Now, let $f_i(x) = a_i x^3 + b_i x$, $i = 1, 2, 3$. We call a polynomial "good modulo $n$" if the residues of $f(0), f(1), \dots, f(n-1)$ modulo $n$ are all distinct.

If $f(x) = a x^3 + b x$ is not good modulo $2013$, then there exists $x_1 \not\equiv x_2 \pmod{2013}$ such that $f(x_1) \equiv f(x_2) \pmod{2013}$. Suppose $x_1 \not\equiv x_2 \pmod{p_i}$. Let $u_1$ and $u_2$ be the residues of $x_1$ and $x_2$ modulo $p_i$, respectively. Then $u_1 \not\equiv u_2 \pmod{p_i}$ and $f_i(u_1) \equiv f_i(u_2) \pmod{p_i}$, so $f_i(x)$ is not good modulo $p_i$.

If $f(x) = a x^3 + b x$ is good modulo $2013$, then for every $i$, $f_i(x)$ is good modulo $p_i$. The reason is as follows. For any distinct pair $r_1, r_2 \in \{0, 1, \dots, p_i - 1\}$, there exist $x_1, x_2 \in \{1, 2, \dots, 2013\}$ such that $x_1 \equiv r_1 \pmod{p_i}$ and $x_2 \equiv r_2 \pmod{p_i}$ and $x_2 \equiv x_2 \pmod{\frac{2013}{p_i}}$. Now $f(x_1) \equiv f(x_2) \pmod{\frac{2013}{p_i}}$, but $f(x_1) \not\equiv f(x_2) \pmod{2013}$, so $f(r_1) \not\equiv f(r_2) \pmod{p_i}$.

Hence, we need to determine the number of good polynomials $f_i(x)$ modulo $p_i$.

For $p_1 = 3$, by Fermat's theorem, a good polynomial
$$f_1(x) \equiv a_1 x^3 + b_1 x \equiv (a_1 + b_1) x \pmod{3}$$
is equivalent to say that $a_1 + b_1$ is not divisible by $3$. There are in total six such $f_1(x)$.

For $i = 2, 3$, if $f_i(x)$ is good modulo $p_i$, then for any $u$ and $v \not\equiv 0 \pmod{p_i}$, $f_i(u + v) \not\equiv f_i(u - v) \pmod{p_i}$, i.e.,
$$f_i(u + v) - f_i(u - v) = 2v[a_i(3u^2 + v^2) + b_i]$$
is not divisible by $p_i$. If $a_i \neq 0$, the residues modulo $p_i$ of elements in the sets $A = \{3a_i u^2 \mid u = 0, 1, \dots, \frac{p_i-1}{2}\}$ and $B = \{(-b_i - a_i v^2) \mid v = 1, 2, \dots, \frac{p_i-1}{2}\}$ do not coincide, and $|A| + |B| = p_i$. So $A \cup B$ forms a complete residue system modulo $p_i$. Their sum must be a multiple of $p_i$, i.e.,
$$\sum_{u=0}^{\frac{p_i-1}{2}} 3a_i u^2 + \sum_{v=1}^{\frac{p_i-1}{2}} (-b_i - a_i v^2) \equiv 0 \pmod{p_i}.$$
Now, $1^2 + 2^2 + \dots + \left(\frac{p_i-1}{2}\right)^2 = \frac{1}{6} \cdot \frac{p_i-1}{2} \cdot \frac{p_i+1}{2} \cdot p_i$ is a multiple of $p_i$, so $-\frac{p_i-1}{2} \cdot b_i$ is also a multiple of $p_i$. Hence, $b_i$ is divisible by $p_i$, i.e., exactly one of $a_i$, $b_i$ is $0$.

If $a_i = 0, b_i \neq 0$, then $f_i(x) = b_i x$ is obviously good. There are $p_i - 1$ such good polynomials.

If $a_i \neq 0, b_i = 0$, then $f_i(x) = a_i x^3$. For $p_2 = 11$, by Fermat's theorem, $(x^3)^7 = x^{21} \equiv x \pmod{11}$, so for $x_1 \neq x_2 \pmod{11}$ and $x_1^3 \neq x_2^3 \pmod{11}$, $f_2(x) = a_2 x^3$ is good. There are in total $10$ such polynomials.

For $p_3 = 61$, as $4^3 = 64 \equiv 125 = 5^3 \pmod{61}$, $f_3(x) = a_3 x^3$ cannot be good.

Therefore, the total number that we are looking for is $6 \times (10 + 10) \times 60 = 7200$. $\square$