Consider n workers, whose wages are $ w_1, w_2, ..., w_n. $
The average wage, then, would be $ A = \frac{w_1 + w_2 + ... + w_n}{n} $ (*).
All wages $ w_i $ can also be expressed as their difference from the average: $ w_i = A + w_{d_i} $.
The total wage paid to all workers, then can be expressed in two different ways:
$ \sum_{i=1}^n w_i = \sum_{i=1}^n A + w_{d_i} $
$ nA = \sum_{i=1}^n A + \sum_{i=1}^n w_{d_i} $ (Using (*) on the LHS)
$ nA = nA + \sum_{i=1}^n w_{d_i} $
$ 0 = \sum_{i=1}^n w_{d_i} $
Consider worker j, who makes more money than average, as posited by Weber. Then, $ w_j = A + w_{d_j} $, where $ w_{d_j} > 0 $.
$ \sum_{i=1}^n w_{d_i} = w_{d_1} + w_{d_2} + ... + w_{d_{j-1}} + w_{d_j} + w_{d_{j+1}} + ... + w_{d_n} = 0 $
$ \Rightarrow w_{d_1} + w_{d_2} + ... + w_{d_{j-1}} + w_{d_{j+1}} + ... + w_{d_n} = -w_{d_j} < 0 $ (Since $ w_{d_j} > 0 $)
Then, at least one of $ w_{d_1}, w_{d_2}, ..., w_{d_{j-1}}, w_{d_{j+1}}, ..., w_{d_n} $ must be negative.
In other words, $ \text{If } \exists j \in {1, 2, ..., n} \text{ s.t. } w_{d_j} > 0, \exists k \in {1, 2, ..., n} (k \neq j) \text{ s.t. } w_{d_k} < 0 $
Or, to summarize: If somebody earns above average, another person must earn below average. Thus, Weber's situation cannot occur.
