Two distinct positive integers are called close if their greatest common divisor equals their difference. Show that for any $n$, there exists a set $S$ of $n$ elements such that any two elements of $S$ are close.
Suppose that $( a_{1} ,\dotsc ,a_{n}) \in \mathbb{N}^{n}$ satisfies the condition.
Let $L=\text{lcm}( a_{1} ,\dotsc ,a_{n})$ and $a_{0} =0$.
Let $b_{i} =L+a_{i}$ for each $i\in \{0,\dotsc ,n\}$
Then we can verify that $( b_{0} ,b_{1} ,\dotsc ,b_{n})$ satisfies the condition as follows:
Suppose that $i\neq j$ and $i,j\in \{0,\dotsc ,n\}$.
We have $\gcd( b_{i} ,b_{j}) =\gcd( L+a_{i} ,L+a_{j}) =\gcd( L+a_{i} ,a_{j} -a_{i})$.
$=\gcd( L+a_{i} ,\gcd( a_{i} ,a_{j})) =\gcd( a_{i} ,a_{j}) =|a_{i} -a_{j} |=|b_{i} -b_{j} |$.
Note that $( 1,2)$ satisfies the condition.
So we can construct a desired set by the above algorithm.
$( 1,2)\rightarrow ( 2,3,4)\rightarrow ( 12,14,15,16)\rightarrow ...$
$\blacksquare $