11.4. A set of composite numbers from the set $\{1,2,3,4, \ldots, 2016\}$ is called good if any two numbers in this set do not have common divisors (other than 1). What is the maximum number of numbers that a good set can have?
Answer: 14 numbers.

Example. $2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}, 19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$. All the presented numbers are composite, as they are squares of natural numbers greater than 1. It is also clear that different prime numbers do not have common divisors greater than 1.

Estimate. Suppose a set of composite numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfies the condition of the problem and $a_{1}<a_{2}<\ldots<a_{n}$. Note then that the inequality $p_{n}^{2} \leq a_{n}$ holds, where $p_{n}$ is the $n$-th prime number in ascending order. Let $q_{1}, q_{2}, \ldots, q_{n}$ be the smallest prime numbers in the factorization of the composite numbers $a_{1}, a_{2}, \ldots, a_{n}$. If for some index $k, 1 \leq k<n$, the inequality $p_{n}<q_{k}$ holds, then $p_{n}^{2}<q_{k}^{2} \leq a_{k}<a_{n}$. If there is no such index, then the prime $q_{n}$ is greater than some $n-1$ prime number. This means that $p_{n}^{2} \leq q_{n}^{2} \leq a_{n}$. Therefore, if there is a set of composite numbers $a_{1}, a_{2}, \ldots, a_{15}$ satisfying the condition of the problem $a_{1}<a_{2}<\ldots<a_{15}$, then $2209=47^{2}=p_{15}^{2} \leq a_{15}$. Contradiction.

Comment. Answer - 1 point, example - 2 points, estimate - 3 points. Points are summed for progress.