Let $a_{1}, a_{2}, \ldots, a_{19}$ be the given numbers and suppose on the contrary that the set $S$ of all $\binom{19}{2} = 171$ numbers, which are written on the paper, are all different,
$$d_{1} < d_{2} < \ldots < d_{171}$$
Denote $k$ as the number of even values among the $19$ given numbers, and $t$ as the number of even values in $S$. It is easy to see that for any $d \in S$, there exist $a_{x}, a_{y}$ such that $d = \operatorname{gcd}(a_{x}, a_{y})$; and this number $d$ is even if and only if $a_{x}, a_{y}$ are both even. Hence, we have $t = \binom{k}{2}$.

Then the number of odd values in $S$ is $171 - t$. Since $d_{171} - d_{1} < 180$ then the number of even values and odd values in $S$ does not exceed $90$, which implies that
$$\left\{ \begin{array}{l} t \leq 90 \\ 171 - t \leq 90 \end{array} \text{ so } 81 \leq \binom{k}{2} \leq 90 .\right.$$
Note that
$$\binom{13}{2} = 78 < 91 = \binom{14}{2}$$
so there does not exist positive integer $k$ such that $81 \leq \binom{k}{2} \leq 90$, contradiction. Hence, the numbers written on the paper cannot be all different.