Hint:
First notice, that when $x$ is an integer, the inequality does not hold.
Let's write $x=n+\alpha$, where $n$ is an integer and $0 < \alpha < 1$, then we can rewrite the inequality as $(n+\alpha)^2 < n(n+1)$. Now the problem is reduced to solving the following inequality:
$$\alpha^2 + 2n \alpha - n < 0$$
Can you take it from here?
First, split your summation of the $2$ terms into $2$ separate summations, i.e.,
$$\sum_{k=0}^m\left(\left\lfloor\frac{n}{m}k\right\rfloor - \left\lceil\frac{n}{m}(k-1)\right\rceil\right) = \sum_{k=0}^m\left\lfloor\frac{nk}{m}\right\rfloor - \sum_{k=0}^m\left\lceil\frac{n(k-1)}{m}\right\rceil \tag{1}\label{eq1A}$$
As suggested in zkutch's comment, we can use several formulas in Wikipedia's Floor and ceiling functions article. First, though, define
$$d = \gcd(m,n), \; \; m = dm_1, \; \; n = dn_1, \; \; \gcd(m_1,n_1) = 1 \tag{2}\label{eq2A}$$
For the first term on the right in \eqref{eq1A}, we have
$$\sum_{k=0}^m\left\lfloor\frac{nk}{m}\right\rfloor = 0 + \sum_{k=1}^{m-1}\left\lfloor\frac{nk}{m}\right\rfloor + n \tag{3}\label{eq3A}$$
For \eqref{eq1A}'s second term on the RHS, using that $\lceil -x \rceil = -\lfloor x \rfloor$, we get
$$\begin{equation}\begin{aligned}
\sum_{k=0}^m\left\lceil\frac{n(k-1)}{m}\right\rceil & = \left\lceil-\frac{n}{m}\right\rceil + 0 + \sum_{k=2}^m\left\lceil\frac{n(k-1)}{m}\right\rceil \\
& = -\left\lfloor\frac{n}{m}\right\rfloor + \sum_{k=1}^{m-1}\left\lceil\frac{nk}{m}\right\rceil
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Using that $\lceil x \rceil = \lfloor x \rfloor + r$, where $r = 0$ if $x$ is an integer, and $r = 1$ otherwise. Using \eqref{eq2A}, we have $\frac{nk}{m} = \frac{dn_{1}k}{dm_{1}} = \frac{n_{1}k}{m_{1}}$. Since $\gcd(m_1,n_1) = 1$, this is an integer only when $k$ is an integral multiple of $m_1$, i.e., $k = m_1, 2m_1, \ldots, (d-1)m_1$. This comprises $d-1$ values among the $m-1$ values being summed. We therefore have
$$\sum_{k=1}^{m-1}\left\lceil\frac{nk}{m}\right\rceil = \sum_{k=1}^{m-1}\left\lfloor\frac{nk}{m}\right\rfloor + (m - 1) - (d - 1) \tag{5}\label{eq5A}$$
Using \eqref{eq5A} in \eqref{eq4A}, with that result along with \eqref{eq3A} in \eqref{eq1A}, gives
$$\begin{equation}\begin{aligned}
& \sum_{k=0}^m\left(\left\lfloor\frac{n}{m}k\right\rfloor - \left\lceil\frac{n}{m}(k-1)\right\rceil\right) \\
& = \left(\sum_{k=1}^{m-1}\left\lfloor\frac{nk}{m}\right\rfloor + n\right) - \left(-\left\lfloor\frac{n}{m}\right\rfloor + \sum_{k=1}^{m-1}\left\lfloor\frac{nk}{m}\right\rfloor + m - d\right) \\
& = n - m + \gcd(m,n) + \left\lfloor\frac{n}{m}\right\rfloor
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Best Answer
Yes, it is true.
$$ \left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor \tag1$$
In the following, $m,n$ are integers.
Case 1 : If $a=2m,b=2n$, then both sides of $(1)$ equal $|m-n|$.
Case 2 : If $a=2m,b=2n+1$, then $$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$
If $m-n-\frac 12\ge 0$, then $m-n-1\ge 0$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.
If $m-n-\frac 12\lt 0$, then $m-n-1\lt 0$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.
Case 3 : If $a=2m+1, b=2n$, then $$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$
If $m-n+\frac 12\ge 0$, then $m-n+1\ge 0$, so$$(3)\iff m-n+1\ge m-n$$which is true.
If $m-n+\frac 12\lt 0$, then $m-n+1\lt 0$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.
Case 4 : If $a=2m+1,b=2n+1$, then both sides of $(1)$ equal $|m-n|$.