Problem 1339

1. Partitioning the Set:
We start by partitioning the set $\{1, 2, \ldots, 4014\}$ into 2007 parts $P_1, P_2, \ldots, P_{2007}$ such that $P_a$ contains all numbers of the form $2^b(2a-1)$ where $b$ is a nonnegative integer. This ensures that no two elements in the same part can divide each other, as one would be a multiple of the other by a power of 2.

2. Choosing Elements from Each Part:
Since $A$ cannot have two elements from the same part, $A$ must have exactly one element from each part. Let $t_a$ be the element of $A$ contained in $P_a$.

3. Analyzing the Sequence:
Consider the elements $t_1, t_2, t_3, \ldots, t_{2007}$. Each $t_a$ is of the form $2^{b_a}(2a-1)$. The highest power of 2 dividing $t_1$ must be strictly greater than the highest power of 2 dividing $t_2$, and so on. This is because if $2^{b_x}(2x-1)$ divides $2^{b_y}(2y-1)$, then $b_x \leq b_y$ and $(2x-1) \mid (2y-1)$.

4. Ensuring Strictly Decreasing Sequence:
The highest powers of 2 dividing $t_1, t_2, t_3, \ldots, t_{2007}$ must form a strictly decreasing sequence. Since there are 2007 elements, the smallest element $t_1$ must be at least $2^{2006}$.

5. Finding the Minimum Element:
To find the minimum $m_A$, we need to ensure that $t_1$ is as small as possible while still satisfying the conditions. The smallest possible $t_1$ is $2^0(2 \cdot 1 - 1) = 1$, but this is not allowed as it would divide any other element. The next smallest possible element is $2^0(2 \cdot 2 - 1) = 3$, but this also does not work as it would divide elements like $6, 12, \ldots$.

6. Ensuring No Divisibility:
We need to find the smallest $t_1$ such that no $t_a$ divides any other $t_b$. By considering the powers of 2 and the odd factors, we find that the smallest possible $t_1$ that satisfies all conditions is $2^7 = 128$.

7. Verification:
We verify that $128$ is indeed the smallest element that can be chosen such that no element in $A$ divides another. This is done by ensuring that for any $t_a = 2^{b_a}(2a-1)$, the sequence of powers of 2 is strictly decreasing and no odd factor $(2a-1)$ divides another.

Therefore, the minimum $m_A$ is $\boxed{128}$.