A

At the highest point of the circular motion for a particle on a string, the tension must counteract the gravitational force. The centripetal acceleration required to keep the particle moving in a circle is given by \( a_c = \frac{v^2}{l} \) where \(v\) is the speed of the particle at the highest point and \(l\) is the length of the string. For the tension to be zero, the necessary centripetal force would solely be due to the weight of the particle. Hence, we must have the following equation at the highest point:
\( mg = \frac{mv^2}{l} \)
This implies that:
\(v^2 = gl\)
Therefore, for the particle to just lose tension at the highest point, the velocity \(u\) at the lowest point must be less than or equal to \(\sqrt{gl}\). Thus, the correct answer is indeed option A.