Confusion with interpreting this power supply's power ratings
Let's say The power supply output is always kept at its max that is to say at 24V and 6.25A. Now if we vary the AC line voltage between 200V up to 240V (by keeping the load current at 6.25A), would the AC input current remain as 1.8A?


No: to see why it is so, let's remember that the apparent input power is related to the load power by the following formula
$$
A_{in}=\frac{P_{out}}{\eta \cos\varphi}\label{1}\tag{1}
$$
where


\$\eta\$ is the efficiency of the power supply
\$\cos\varphi\$ is the power factor.


Now, assuming that \$\eta, \cos\varphi\$ are approximatively constant respect to the output power \$P_{out}\$, and remembering that
$$
A_{in}=V_{\mathrm{AC_{in}}}\cdot I_{\mathrm{AC_{in}}},
$$ 
by substituting this equation in formula \eqref{1}, we get
$$
I_{\mathrm{AC_{in}}}\simeq \frac{P_{out}}{V_{\mathrm{AC_{in}}}\eta \cos\varphi }\label{2}\tag{2}
$$
therefore, keeping the output power constant and lowering the input voltage makes the input current rising.

Few observations


The nominal input current (\$1.8\mathrm{A}\$ for the range considered in the question) for a given range usually refers to the maximum value measured at the minimum input voltage (if the datasheet is properly written). This is due to the fact that this value also serves as a design parameter for the upstream part of the circuit (overload protection devices like fuses and other kind of breakers), so the designer should rely on it.
The assumptions \$\eta\simeq\mathrm{const.}\$ and \$\cos\varphi\simeq\mathrm{const.}\$ are only approximately true, thus you can expect noticeable variations in the real behavior \$I_{\mathrm{AC_{in}}}=f(V_{\mathrm{AC_{in}}})\$ respect to the one predicted by formula \eqref{2}. However, ceteris paribus, you can always expect an increase of the input current when lowering the input voltage.