Question. How can we prove that this function is non-decreasing?
That is:
$$ f: \Bbb{R} \to \Bbb{Z} \\ f(x) = [\frac{x}{1}] - [\frac{x-1}{2}] - [\frac{x - 1}{3}] - [\frac{x - 2}{3}] - [\frac{x - 1}{5}] - [\frac{x - 4}{5}] \\ + [\frac{x - 1}{6}] + [\frac{x - 5}{6}] + [\frac{x-1}{10}] + [\frac{x - 9}{10}] + [\frac{x - 1}{15}] + [\frac{x-4}{15}] + [\frac{x - 11}{15}] \\ - [\frac{x - 1}{30}] - [\frac{x - 11}{30}] - [\frac{x - 19}{30}] - [\frac{x - 29}{30}] $$
or more compactly:
$$ f(x) = \sum_{d \mid p_3\#}(-1)^{\omega(d)}\sum_{0 \leq r \lt d \\ r^2 = 1 \pmod d}\left\lfloor \frac{x - r}{d}\right\rfloor $$
By this very elegant MSE answer: obviously there is a trend line, and the slope of this trend line can be computed as:
$$ m = \sum_{d\mid p_3\#} \frac{(-1)^{\omega(d)}|H_d|}{d} $$
where $H_d = \{r \in \Bbb{Z} : 0 \leq r \lt d \ \wedge \ r^2 = 1 \pmod d\}$. It is known that $H_d$ is precisely the set of standard residues of a subgroup of $(\Bbb{Z}/d)^{\times}$, namely the subgroup of all $x \in \Bbb{Z}/d$ that square to $1$ modulo $d$.
A more general number-theoretic answer:
Note that $f(x)$ only changes at integers. $$f(n + 1) - f(n) = \sum_{d | p_3 \#} {(-1)^{\omega(d)} \sum_{0\leq r< d\\r^2\equiv 1\pmod d} \lfloor\frac{n+1-r}d\rfloor - \lfloor\frac{n-r}d\rfloor} = \sum_{d | p_3 \#} {(-1)^{\omega(d)} \sum_{0\leq r< d\\r^2\equiv 1\pmod d} [n+1\equiv r \pmod d]} = \sum_{d | p_3 \#} {(-1)^{\omega(d)} [(n+1)^2\equiv 1 \pmod d]} = \sum_{d | p_3 \#} {(-1)^{\omega(d)} \prod_{p | d}[(n+1)^2\equiv 1 \pmod p]} = \sum_{d | p_3 \#} {\prod_{p | d}- [(n+1)^2\equiv 1 \pmod p]} = \prod_{p | p_3\#} {1 - [(n+1)^2\equiv 1 \pmod p]}$$
Which is obviously nonnegative.
Old solution:
Note that $\lfloor \frac{x - r}{b} \rfloor = \frac{x - r - (x-r\mod b)}{b} = \frac{x-r-((x\mod b) - (r\mod b) \mod b)}{b}$, so $\lfloor \frac{x - r}{b} \rfloor - \frac{x}{b}$ is periodic with period $b$. Because $f(x) - mx$ is a sum of terms each with a period dividing $30$, $f(x) - mx$ is periodic with period $30$.
We'll show that if $f(x)$ is non-decreasing for $0 \leq x < 60$, then it must be non-decreasing in general.
Suppose $x < y$, and we'll show $f(x) \leq f(y)$. Due to the periodicity $f(x) \leq f(y)$ iff $f(x + 30k) \leq f(y + 30k)$, so we can assume $0 \leq x < 30$.
We will use induction on $\lfloor \frac y{30} \rfloor$. As a base case, if $y < 60$ then $f(x) \leq f(y)$ follows by our assumption of $f$ not decreasing in the range $0 \leq x < 60$. Otherwise $x < y - 30$, and by induction $f(x) \leq f(y - 30)$, but due to the periodicity $f(y - 30) = f(y) - 30m < f(y)$, so $f(x) < f(y)$.
To show it's non-decreasing in the range, since the function only changes at integers, we can just evaluate it for the entire range, or look at the graph.