Let this subset be $\{a_1, a_2, \cdots, a_{27}\}.$ For $1 \le i \le 26,$ let $b_i = \gcd(a_i, a_{27}).$ Notice that all of the $b_i$'s must be distinct and that they all must divide $a_{27}.$ Indeed, the $b_i$'s are distinct because $b_i = b_j \Rightarrow \gcd(a_i, a_{27}) | a_j.$ Similarly, we have not only that the $b_i$'s are distinct, but that no two of them divide one another. Let $a_{27} = p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}$ be the prime factorization of $a_{27}.$ Observe that no two of the $b_i$'s can have the same $v_{p_2}, v_{p_3}, \cdots, v_{p_k}$, because that would imply that one divides the other. Therefore, we conclude that there are at least $26 \cdot (k_1 + 1) \ge 52$ divisors of $a_{27}.$ However, it's easy to check that all positive integers at most $2007$ have at most $40$ divisors ($1680$ has $40$). To check this, first easily obtain that $60$ divides any number in $1, 2, \cdots, 2007$ with a maximal number of prime divisors, and that any such number must be $>1000$ (else double it). This reduces our search space to $17$ numbers. $\square$