Let $A = p_1^{k_1} p_2^{k_2} \dots p_n^{k_n}$ be a canonical prime factorization of $A$, where $p_i$ are different prime factors, and $k_i$ are positive integers, $i = 1, n$. Knowing the total number of positive integer factors of $A$ we can conclude that
$$(k_1 + 1)(k_2 + 1)\dots(k_n + 1) = 2018.$$

Since the left-hand side comprises factors each of which is greater or equal to 2, we have only two cases:

1) $k_1 + 1 = 2018$.
2) $(k_1 + 1)(k_2 + 1) = 2 \cdot 1009$.

From the first case, $A$ has a canonical prime factorization of the form: $A = p_1^{2017}$, which contradicts the fact that $A \neq 2018$, since in that case $A$ can't be divisible by both 2 and the prime number 1009.

Hence, $(k_1 + 1)(k_2 + 1) = 2 \cdot 1009$, which yields, without loss of generality, $k_1 = 1$, $k_2 = 1008$. Therefore, $A = p_1 p_2^{1008}$, moreover from divisibility by 2 and 1009 we obtain $A = 2 \cdot 1009^{1008}$ or $A = 1009 \cdot 2^{1008}$. However, none of these numbers is divisible by $2018^2$, Q.E.D.