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.