Note that both $\frac n3-\left\lfloor\frac n3\right\rfloor$ and $\left\lceil\frac{2n}3\right\rceil-\frac{2n}3$ have a period of $3$. Subtracting we get that
$$
\left\lceil\frac{2n}3\right\rceil+\left\lfloor\frac n3\right\rfloor-n\tag{1}
$$
has a period of $3$. Thus, we only need to check $n=0,1,2$ to verify
$$
\left\lceil\frac{2n}3\right\rceil+\left\lfloor\frac n3\right\rfloor=n\tag{2}
$$
for all $n$.
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 $|m-n|$.
Best Answer
Let $m = \lceil \log_3(x) \rceil$ and $n = \lfloor \log_2(x) \rfloor$. Thus $m$ and $n$ are integers, $3^m \ge x > 3^{m-1}$ and $2^{n+1} > x \ge 2^n$. If $n < m$ we'd have $n \le m-1$, and $3^{m-1} < x < 2^{n+1} \le 2^m$, so $(3/2)^m = 3^m/2^m < 3 $. Since $(3/2)^3 = 27/8 > 3$, $m < 3$ and $n < 2$.
Now for $x = 1$, $\lceil \log_3(1) \rceil = 0 = \lfloor \log_2(1) \rfloor$; for $x = 2$, $\lceil \log_3(2) = 1 = \lfloor \log_2(2) \rfloor$; for $x = 3$, $\lceil \log_3(3) = 1 = \lfloor \log_2(3) \rfloor$; for $x \ge 4$, $\lfloor \log_2(x) \rfloor \ge 2$.
So that leaves no possible positive integers $x$ for which we could have $n < m$.