Sorry, you are right. And here is the proof of the correctness.

Let x_i be the number on player i 's head, y_i be his output, e_i be his offset and f_i be his error. We would like to show that there exists some i s.t. f_i = 0.

For every i, we have (all arithmetic modulo n)
y_i = e_i - (x_1+...+x_n -x_i) = x_i - f_i.
Therefore e_i+f_i=x_1+...+x_n. Since {e_i} is the set of congruence modulo n, it follows that {f_i} is also {0,1,...,n-1}; in particular, there exist some j with f_j=0.

Very nice argument.