Let's say, $q\in {]}0,100{]}$ is the minimal net percentage of the volume (or of the mass) that evaporates at any second $t$ (for every $t>0$). Saying "net", we assume that more water molecules leave the kitchen counter than return, and that the fraction of the molecules leaving the surface in relation to the number of molecules getting back has a constant, positive lower bound. (Other answers explain why it is likely to be so in kitchen conditions.)
Then, at most $100-q$ percent are left per time unit. So, after $t$ time units, the amount of water left will be at most $\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t$, where $a_0$ is the initial amount. Since $100-q<100$, we obtain $$\lim_{t\to+\infty}\,\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t \ = \ 0\,.$$ In particular, after a certain point of time, the amount of water will be lower than the minimal possible amount (the volume of one $\mathrm{H}_2\mathrm{O}$ molecule or its mass, simplified, of course).
If the assumption made is not valid (say, due to great humidity somewhere in Asia), the result would be wrong: the water will NOT fully evaporate.
(An aside has to be made. Note that the above mathematical treatment is a gross simplification. To get a more realistic evaporation model, we should take into account that the evaporation happens only from the surface, and not from the whole volume, and that both the surface and the volume change with time. Also, bear in mind that even within one second, the evaporation rate changes.)