Solution: First, point out two obvious conclusions:
(1) If $m \in \mathbf{Z}_{+}, x \in \mathbf{R}$, then $\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right]$;
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$

In (1), let $m=k!(k=1,2, \cdots, 2013)$ and sum up to get
$$
f(x)=\sum_{k=1}^{2013}\left[\frac{x}{k!}\right] = \sum_{k=1}^{2013}\left[\frac{[x]}{k!}\right]=f([x]) .
$$

This shows that the equation $f(x)=n$ has a real solution if and only if it has an integer solution. Therefore, we only need to consider the case where $x$ is an integer.
$$
\begin{array}{l}
\text { By } f(x+1)-f(x) \\
=[x+1]-[x]+\sum_{k=2}^{2013}\left(\left[\frac{x+1}{k!}\right]-\left[\frac{x}{k!}\right]\right) \\
\geqslant 1,
\end{array}
$$

we know that $f(x)(x \in \mathbf{Z})$ is strictly monotonically increasing.
Next, find integers $a$ and $b$ such that
$$
\begin{array}{l}
f(a-1)2013,
\end{array}
$$

so $b=1173$.
Thus, the good numbers in $\{1,3, \cdots, 2013\}$ are the odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$.
In (1), let $x=2 l(l=0,1, \cdots, 586)$.
By conclusion (2), we know
$\left[\frac{2 l+1}{k!}\right]=\left[\frac{2 l}{k!}\right](2 \leqslant k \leqslant 2013)$.
Then $f(2 l+1)-f(2 l)$
$$
=1+\sum_{k=2}^{2013}\left(\left[\frac{2 l+1}{k!}\right]-\left[\frac{2 l}{k!}\right]\right)=1 \text {. }
$$

This shows that exactly one of $f(2 l)$ and $f(2 l+1)$ is odd. Therefore, there are $\frac{1174}{2}=587$ odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$, meaning there are 587 good numbers in the set $\{1,3, \cdots, 2013\}$.

[Note] In this problem, converting the count of good numbers in $\{1,3, \cdots, 2013\}$ to the count of odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$ is highly valuable. Furthermore, pairing $f(2 l)$ and $f(2 l+1)$ and finding that exactly one in each pair is odd makes the relatively complex task of determining the parity of $f(2 l)$ and $f(2 l+1)$ for each $l$ unnecessary.
Example 4 Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the set $\{1,3, \cdots, 2013\}$. ${ }^{\text {[4] }}$
(2013, China Southeast Mathematical Olympiad)