If $n \in A$ and $n = p^2$, then $1000 \le p^2 \le 2014$, that is $32 \le p \le 44$. The largest subset of $A$ whose elements are perfect squares is
$$B = \{32^2, 33^2, 34^2, 35^2, 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, 44^2\}.$$
We must choose among them the maximum number of pairwise prime numbers. Consider the partition of $B$ into the sets $C_1 = \{32^2, 34^2, 36^2, 38^2, 40^2, 42^2, 44^2\}$, $C_2 = \{33^2, 39^2\}$, $C_3 = \{35^2\}$, $C_4 = \{37^2\}$, $C_5 = \{41^2\}$, $C_6 = \{43^2\}$. If we choose 7 or more elements of $B$, then two of them are in the same $C_i$, so they are not co-prime. So we cannot take more than 6 elements; an example is $\{32^2, 33^2, 35^2, 37^2, 41^2, 43^2\}$.