China Southeastern Mathematical Olympiad

First, we point out two obvious facts:
(a) If $m$ is a positive integer and $x$ is real, then
$$\lfloor \frac{x}{m} \rfloor = \lfloor \frac{\lfloor x \rfloor}{m} \rfloor.$$
(b) For any integer $l$ and positive even number $m$, we have
$$\lfloor \frac{2l + 1}{m} \rfloor = \lfloor \frac{2l}{m} \rfloor.$$
Let $m = k!$ ($k = 1, 2, \dots, 2013$) in (a) and summing up, we have
$$f(x) = \sum_{k=1}^{2013} \lfloor \frac{x}{k!} \rfloor = \sum_{k=1}^{2013} \lfloor \frac{\lfloor x \rfloor}{k!} \rfloor = f(\lfloor x \rfloor),$$
that is, $f(x) = n$ has a real solution if and only if $f(x) = n$ has an integer solution. So, we only consider $x$ as an integer. Since
$$f(x+1) - f(x) = \lfloor x+1 \rfloor - \lfloor x \rfloor + \sum_{k=2}^{2013} \left( \lfloor \frac{x+1}{k!} \rfloor - \lfloor \frac{x}{k!} \rfloor \right) \ge 1, \quad \textcircled{1}$$
we see that $f(x)$ ($x \in \mathbb{Z}$) is monotonously increasing. Now, we find integers $a$ and $b$, such that
$$\begin{align*} f(a-1) < 0 &\le f(a) < f(a+1) < \dots \\ &< f(b-1) < f(b) \le 2013 < f(b+1). \end{align*}$$
Note that $f(-1) < 0 = f(0)$, so $a = 0$. Since
$$\begin{align*} f(1173) &= \sum_{k=1}^{6} \left\lfloor \frac{1173}{k!} \right\rfloor = 1173 + 586 + 195 + 48 + 9 + 1 \\ &= 2012 \le 2013, \end{align*}$$
$$\begin{align*} f(1174) &= \sum_{k=1}^{6} \left\lfloor \frac{1174}{k!} \right\rfloor = 1174 + 587 + 195 + 48 + 9 + 1 \\ &= 2014 > 2013, \end{align*}$$
we see that $b = 1173$.
So the good numbers in $\{1, 3, 5, ..., 2013\}$ are the odd numbers in
$$\{f(0), f(1), \dots, f(1173)\}.$$
Let $x = 2l$ ($l = 0, 1, \dots, 586$) in ①. By (b), we have
$$\left\lfloor \frac{2l + 1}{k!} \right\rfloor = \left\lfloor \frac{2l}{k!} \right\rfloor \quad (2 \le k \le 2013).$$
Thus,
$$f(2l + 1) - f(2l) = 1 + \sum_{k=2}^{2013} \left( \left\lfloor \frac{2l+1}{k!} \right\rfloor - \left\lfloor \frac{2l}{k!} \right\rfloor \right) = 1,$$
that is, there is exactly one odd number in $f(2l)$ and $f(2l + 1)$.
Therefore, there are $\frac{1174}{2} = 587$ odd numbers in $\{f(0), f(1), \dots, f(1173)\}$, that is, there are 587 good numbers in the set $\{1, 3, 5, \dots, 2013\}$.
$\square$