For any $k \in \{1, 2, ..., n\}$ denote by $p_k$ the number of visitors in the $k$-th row. The stated condition on given $i$ and $j$ implies that the number of friendly pairs $(A, B)$, where $A$ and $B$ are from the $i$-th row and from $j$-th row respectively, is equal to the product $j p_i$. Interchanging the indices $i$ and $j$, we conclude that the same number of friendly pairs $(A, B)$ equals $i p_j$. Thus $j p_i = i p_j$ or $p_i : p_j = i : j$, and therefore, all the numbers $p_k$ must be proportional as follows:
$$p_1 : p_2 : \dots : p_n = 1 : 2 : \dots : n.$$
Let us show that under this proportionality the visitors can be friendly in such a way which ensures the property under consideration. Thus assume that for some positive integer $d$, the equality $p_k = k d$ holds with any $k \in \{1, 2, ..., n\}$. Let us

start with the case $d = 1$ when the numbers of visitors in single rows are successively $1, 2, ..., n$. Then the stated property holds true if (and only if) any two visitors — taken from distinct rows in the whole auditorium — are friends. In the case when $d > 1$, let us divide all the visitors into $d$ groups $G_1, G_2, ..., G_d$ so that for arbitrary $k = 1, 2, ..., d$, the numbers of visitors from the group $G_k$ in single rows are successively $1, 2, ..., n$. It is evident that the stated property holds true under the following condition: two visitors are friends if and only if they belong to the same group $G_k$.
It follows from the preceding that our task is to find such integer values of $n$, $n \ge 4$, for which there exists a positive integer $d$ satisfying the equation
$$d + 2d + \dots + n d = 234 \quad \text{or} \quad d n(n+1) = 468.$$
Thus we look for all divisors $468 = 2^2 \cdot 3^2 \cdot 13$ which are of the form $n(n+1)$. Inequality $22 \cdot 23 > 468$ implies that $n < 22$ and hence $n \in \{4, 6, 9, 12, 13, 18\}$. It is easy to see that the only satisfactory $n$ equals $12$ (which corresponds to $d=3$).

Answer. The unique solution is $n = 12$.