The average of 25 numbers is zero. At most, how many of these numbers can be greater than zero ?

Let's assume the 25 numbers are $$N_{1}, N_{2}, N_{3}, N_{4}, N_{5} .................... N_{24} N_{25}$$.

given that average of 25 numbers is 0

$$\frac{\left(N_1+N_2+N_3+N_4+N_5....................\ +N_{24}+N_{25}\right)}{25}\ =\ 0$$

$$N_1+N_2+N_3+N_4+N_5.................... +N_{24}+N_{25} = 0$$

$$N_1+N_2+N_3+N_4+N_5.................... +N_{24} = - N_{25}$$

From the above equation, we can say that maximum 24 numbers will be greater than zero and one number will be negative.

OR

we can assume that the sum of 24 number $$N_1+N_2+N_3+N_4+N_5.................... +N_{24}$$ will be 1 and $$N_{25}$$ will be -1. So the average of these will be zero.