Maximum possible value of a positive integer $n$, such that for any choice of seven distinct elements from ${1, 2, .., n},$ there will exist What is the maximum possible value of a positive integer $n$, such that for any choice of seven distinct elements from ${1, 2, ..., n},$ there will exist two numbers $x$  &  $y$ satisfying $1 < x/y \leq 2$
What seems unclear to me are the following points,
A) Is $2y$ supposed to be $\leq$ n ?
B) Is there some standard approach to find the maximum possible value?
Moreover, usually, I try to take a small set to try out the given conditions. In this case, is it possible to take a smaller set and use it to draw conclusions for the larger set?
The answer is $2^7 - 2$

Starting with smaller values is the key here. Elimination can be done once a pattern is seen.
On the basis of the conditions given ($1 < x/y \leq 2$), try taking some numbers. Notice that, as mentioned in the comments, Choosing $1$ restricts the entry of $2\times1$
Thereafter, choosing $3$, restricts the entry of all numbers $\leq 2 \times 3$
Finally we are left with $7$. This clearly points to the fact/pattern that we get a number that is of the form $2^t-1$. Continuing the trend, we easily observe that the set $\{1,3,7,15,...,127 = 2^7-1\}$ is a set consisting of seven elements, no two $x,y$ of which satisfy the criteria that $1 < \frac{x}{y} \leq 2$.  Therefore, we know that the answer to the question is at most $127$.
We guess that it is $126$. Indeed, here is a proof. Divide $126$ into the following parts : $S_1 =\{1,2\}$, $S_2 = \{3,4,5,6\}$, $S_3 = \{7,8,...,14\}$, ..., $S_6 = \{63,64,...,126\}$. Now, if you pick any two elements $x,y$ from the same $S_i$, then either $1 < \frac{x}{y} \leq 2$ or $1 < \frac {y}{x} \leq 2$. This can be easily checked.
By the pigeonhole principle, if we pick $7$ elements between $1$ and $126$, two of them must be in the same $S_i$, and hence we are done.
