4. Find the number of polynomials $f(x)=a x^{3}+b x$ that satisfy the following two conditions:
(1) $a, b \in\{1,2, \cdots, 2013\}$;
(2) The difference between any two numbers in $f(1), f(2), \cdots, f(2013)$ is not a multiple of 2013. (Wang Bin)
4. It is known that the prime factorization of $2013=3 \times 11 \times 61$.

Let $p_{1}=3, p_{2}=11, p_{3}=61$.
For $a, b \in\{1,2, \cdots, 2013\}$, let
$a \equiv a_{i}\left(\bmod p_{i}\right), b \equiv b_{i}\left(\bmod p_{i}\right)$,
where $i=1,2,3$.
By the Chinese Remainder Theorem, we know that $(a, b)$ and $\left(a_{1}{ }^{\prime}, a_{2}, a_{3}\right.$, $\left.b_{1}, b_{2}, b_{3}\right)$ are in one-to-one correspondence.
Let $f_{i}(x)=a_{i} x^{3}+b_{i} x(i=1,2,3)$.
A polynomial is called a "good polynomial modulo $n$" if $f(0), f(1), \cdots, f(n-1)$ have distinct remainders modulo $n$.

If $f(x)=a x^{3}+b x$ is not a good polynomial modulo 2013, then there exist $x_{1} \neq x_{2}(\bmod 2013)$ such that
$f\left(x_{1}\right) \equiv f\left(x_{2}\right)(\bmod 2013)$.
Assume $x_{1} \neq x_{2}\left(\bmod p_{i}\right)$, and let $u_{1}, u_{2}$ be the remainders of $x_{1}$ and $x_{2}$ modulo $p_{i}$. Then
$u_{1} \equiv u_{2}\left(\bmod p_{i}\right)$,

and $f_{i}\left(u_{1}\right) \equiv f_{i}\left(u_{2}\right)\left(\bmod p_{i}\right)$.
Thus, $f_{i}(x)$ is not a good polynomial modulo $p_{i}$.
If $f(x)=a x^{3}+b x$ is a good polynomial modulo 2013, then each $f_{i}(x)$ is a good polynomial modulo $p_{i}$.
In fact, for any different
$r_{1}, r_{2} \in\left\{0,1, \cdots, p_{i}-1\right\}$,

there exist $x_{1}, x_{2} \in\{1,2, \cdots, 2013\}$ such that
$$
x_{1} \equiv x_{2}\left(\bmod \frac{2013}{p_{i}}\right)
$$

and $x_{1} \equiv r_{1}\left(\bmod p_{i}\right), x_{2} \equiv r_{2}\left(\bmod p_{i}\right)$.
Since $f\left(x_{1}\right) \neq f\left(x_{2}\right)(\bmod 2013)$ and
$f\left(x_{1}\right) \equiv f\left(x_{2}\right)\left(\bmod \frac{2013}{p_{i}}\right)$,
it follows that $f\left(r_{1}\right) \neq f\left(r_{2}\right)\left(\bmod p_{i}\right)$, and the conclusion holds.
Therefore, the problem reduces to finding the number of good polynomials $f_{i}(x)$ modulo $p_{i}$.
For $p_{1}=3$, by Fermat's Little Theorem, we have
$f_{1}(x) \equiv a_{1} x+b_{1} x \equiv\left(a_{1}+b_{1}\right) x(\bmod 3)$

is a good polynomial if and only if $a_{1}+b_{1}$ is not a multiple of 3, and there are 6 such $f_{1}(x)$.
For $i=2,3$, if $f_{i}(x)$ is a good polynomial modulo $p_{i}$,

then for any $u, v \not\equiv 0\left(\bmod p_{i}\right)$,
$f_{i}(u+v) \not\equiv f_{i}(u-v)\left(\bmod p_{i}\right)$
$\Rightarrow p_{i} \nmid\left[f_{i}(u+v)-f_{i}(u-v)\right]$
$\Rightarrow p_{i} \nmid 2 v\left[a_{i}\left(3 u^{2}+v^{2}\right)+b_{i}\right]$.
If $a_{i} \neq 0$, the sets
$A=\left\{3 a_{i} u^{2} \mid u=0,1, \cdots, \frac{p_{i}-1}{2}\right\}$

and $B=\left\{-b_{i}-a_{i} v^{2} \mid v=1,2, \cdots, \frac{p_{i}-1}{2}\right\}$,
have non-overlapping remainders modulo $p_{i}$.
Clearly, the elements of sets $A$ and $B$ are distinct modulo $p_{i}$, and $|A|+|B|=p_{i}$, so the $p_{i}$ elements of $A \cup B$ have remainders that are a permutation of $0,1, \cdots, p_{i}-1$, and thus their sum is a multiple of $p_{i}$, i.e.,
$$
\sum_{u=0}^{\frac{p_{i}-1}{2}} 3 a_{i} u^{2}+\sum_{v=1}^{\frac{p_{i}-1}{2}}\left(-b_{i}-a_{i} v^{2}\right) \equiv 0\left(\bmod p_{i}\right) .
$$

Notice that,
$$
\begin{array}{l}
1^{2}+2^{2}+\cdots+\left(\frac{p_{i}-1}{2}\right)^{2} \\
=\frac{1}{6} \times \frac{p_{i}-1}{2} \times \frac{p_{i}+1}{2} p_{i}
\end{array}
$$

is a multiple of $p_{i}$, so $-\frac{p_{i}-1}{2} b_{i}$ is also a multiple of $p_{i}$. Thus, $b_{i}$ is a multiple of $p_{i}$, meaning that at least one of $a_{i}$ or $b_{i}$ is 0, and they cannot both be 0.

If $a_{i}=0, b_{i} \neq 0$, then $f_{i}(x)=b_{i} x$ is clearly a good polynomial.
Thus, 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 Little Theorem, we have
$\left(x^{3}\right)^{?}=x^{21} \equiv x(\bmod 11)$.
Thus, for $x_{1} \equiv x_{2}(\bmod 11), x_{1}^{3} \equiv x_{2}^{3}(\bmod 11)$, $f_{2}(x)=a_{2} x^{3}$ is a good polynomial.
Thus, there are 10 such good polynomials.
For $p_{3}=61$, since
$4^{3}=64 \equiv 125=5^{3}(\bmod 61)$,
$f_{3}(x)=a_{3} x^{3}$ is not a good polynomial.
In summary, the number of polynomials $f(x)$ is
$6 \times(10+10) \times 60=7200$.