Brazilian Math Olympiad

The answer is yes and you can construct an example in several ways. The main observation is that $\gcd(a_i, a_j) = a_j - a_i \iff a_j - a_i \mid a_i$. In fact, if $\gcd(a_i, a_j) = a_j - a_i$ then $a_j - a_i \mid a_i$ and, conversely, if $a_j - a_i \mid a_i$ then $a_j - a_i \mid a_i + (a_j - a_i) \iff a_j - a_i \mid a_j$, so $a_j - a_i \mid \gcd(a_i, a_j)$. But $\gcd(a_i, a_j) \mid a_i$ and $\gcd(a_i, a_j) \mid a_j$ implies $\gcd(a_i, a_j) \mid a_j - a_i$, so $\gcd(a_i, a_j) = a_j - a_i$.

Once this fact is established, one can construct the sequence inductively as follows: first consider the two-term sequence $(1, 2)$. Now, given a sequence
$(x_1, x_2, \dots, x_{k-1})$ with $k-1$ terms such that $\gcd(x_i, x_j) = x_j - x_i$, construct a new sequence adding $x_0$ to every term and putting $x_0$ at its beginning: $(x_0, x_1 + x_0, x_2 + x_0, \dots, x_{k-1} + x_0)$. All we need to do is to find $x_0$. By the previous observation, we need $x_j - x_i \mid x_i + x_0$ and $x_i \mid x_0$. We already have that $x_j - x_i \mid x_i$, so a good choice is $x_0 = \operatorname{lcm}(x_1, x_2, \dots, x_{k-1})$, because by definition $x_i \mid x_0$ and, since $x_i \mid x_0$ and $x_j - x_i \mid x_i$, $x_j - x_i \mid x_0$, so $x_j - x_i \mid x_i + x_0$. So we obtained a new sequence with $k$ terms and the result follows by induction.