If the average of 100 numbers is zero. Of them, at the most, how many may be greater than zero?

Sure, let's break down the problem step by step.

1. Understanding the Average:

   • The average of 100 numbers is given as zero.
   • The formula for the average of a set of numbers is the sum of the numbers divided by the count of the numbers.
   • Therefore, if the average of 100 numbers is zero, the sum of these 100 numbers must be zero.

2. Sum of the Numbers:

   • Let’s denote the 100 numbers as $a_1, a_2, a_3, \ldots, a_{100}$.
   • The sum of these numbers is $a_1 + a_2 + a_3 + \ldots + a_{100} = 0$.

3. Positive and Negative Numbers:

   • To find the maximum number of positive numbers, we need to consider the balance between positive and negative numbers.
   • If some numbers are positive, their sum must be balanced by an equal sum of negative numbers to ensure the total sum is zero.

4. Maximizing Positive Numbers:

   • To maximize the number of positive numbers, we should minimize the number of negative numbers.
   • The smallest negative number that can balance the sum of positive numbers is a single negative number.

5. Scenario with One Negative Number:

   • Suppose there are $n$ positive numbers and 1 negative number.
   • Let the positive numbers be $p_1, p_2, \ldots, p_n$ and the negative number be $-N$.
   • The sum of the positive numbers is $p_1 + p_2 + \ldots + p_n$.
   • For the total sum to be zero, the sum of the positive numbers must equal the negative number: $p_1 + p_2 + \ldots + p_n = N$.

6. Number of Positive Numbers:

   • Since there are 100 numbers in total and 1 of them is negative, the maximum number of positive numbers is $100 - 1 = 99$.

Therefore, at most, there can be 99 positive numbers among the 100 numbers if their average is zero.