16. Existence.
Below is the proof using mathematical induction.
(1) Since $|a-b| \geqslant (a, b)$, we have $|a-b| = (a, b) \Leftrightarrow (a-b) \mid a$.
Clearly, 2 and 3 satisfy the condition.
(2) Assume there exist $k\left(k \in \mathbf{N}_{+}\right)$ positive integers $a_{1}, a_{2}$,
$\cdots, a_{k}$, such that for any $1 \leqslant i < j \leqslant k$, we have
$$\left|a_{i}-a_{j}\right| \mid a_{i} \text{. }$$

Then we can choose such $k+1$ numbers:
$$\begin{array}{l} b_{1}=a_{1} a_{2} \cdots a_{k}, \\ b_{2}=a_{1} a_{2} \cdots a_{k} + a_{1}, \\ b_{3}=a_{1} a_{2} \cdots a_{k} + a_{2}, \\ \cdots \cdots \cdot \\ b_{k+1}=a_{1} a_{2} \cdots a_{k} + a_{k} . \end{array}$$

For any $1 \leqslant i < j \leqslant k+1$, we have
$$b_{i}-b_{j}=a_{i-1}-a_{j-1}\left(a_{0}=0\right) \text{. }$$

Since $\left(a_{i-1}-a_{j-1}\right) \mid a_{1} a_{2} \cdots a_{k}$,
$$\left(a_{i-1}-a_{j-1}\right) \mid a_{i-1} \text{, }$$

it follows that $\left(a_{i-1}-a_{j-1}\right) \mid b_{i}$.
Thus, $\left(b_{i}-b_{j}\right) \mid b_{i}$.
Therefore, there exist $k+1$ numbers that satisfy the problem's requirements.
From (1) and (2), we can conclude that there exist any finite number of numbers that satisfy the requirement, and of course, 2011 numbers can also be found.