5. 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 $\{1,3,5, \cdots, 2013\}$.
（Wu Genxiu, Contributed）
5. First, give two obvious conclusions:
(1) If $m$ is a positive integer and $x$ is a real number, then
$$
\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right] ;
$$
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$

Returning to the original problem.
In conclusion (1), let $m=k!(k=1,2, \cdots, 2013)$, and summing up, we 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 the equation $f(x)=n$ has an integer solution.
From now on, 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 monotonically increasing.
Next, find integers $a$ and $b$ such that
$$
\begin{array}{l}
f(a-1)2013,
\end{array}
$$

so $b=1173$.
Therefore, the good numbers in $\{1,3,5, \cdots, 2013\}$ are the odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$.

In equation (1), let $x=2 l(l=0,1, \cdots, 586)$, by conclusion (2) we know
$$
\begin{array}{l}
{\left[\frac{2 l+1}{k!}\right]=\left[\frac{2 l}{k!}\right](2 \leqslant k \leqslant 2013) .} \\
\text { Hence } f(2 l+1)-f(2 l) \\
=1+\sum_{k=2}^{2013}\left(\left[\frac{2 l+1}{k!}\right]-\left[\frac{2 l}{k!}\right]\right)=1,
\end{array}
$$

This shows that $f(2 l)$ and $f(2 l+1)$ have exactly one odd number.
Thus, $\{f(0), f(1), \cdots, f(1173)\}$ contains exactly $\frac{1174}{2}=587$ odd numbers, i.e., the set $\{1,3,5, \cdots, 2013\}$ contains 587 good numbers.