What is the $∆H_{vap}$ when temperature of a given liquid is equal to the boiling point of the liquid?
For the other cases ie (i) T>BP, $∆H_{vap}$ > 0 (ii) T<BP, $∆H_{vap}$ < 0
So what will be the $∆H_{vap}$ when T (Temperature) is exactly equal to BP of the liquid?
Because we know that heat has to be an input to convert from liquid phase to gaseous phase so can the the $∆H_{vap}$ = 0 ? If so, how is it possible since we are inputting heat into the container.
$\Delta H_\mathrm{vap} \gt 0$ is valid regardless of the relation $T$ vs $T_\mathrm{vap}$, as it is an endothermic process, requiring energy to break intermolecular bonds. Its value drifts with T, depending on the relation of the specific/molar heat capacity of the liquid and the vapour, following the Hess's Law.
You probably confuse $\Delta H_\mathrm{vap}$ which is always positive and $\Delta G_\mathrm{vap} = \Delta H_\mathrm{vap} - T \cdot \Delta S_\mathrm{vap}$, which is zero for the phase equilibrium for given $T$ or $p$ ( sitting on the (l)-(g) line of the phase diagram.)
If $T \lt T_\mathrm{vap}$ for the given $p$ then $\Delta G_\mathrm{vap} \gt 0$
If $T \gt T_\mathrm{vap}$ for the given $p$ then $\Delta G_\mathrm{vap} \lt 0$
