Let's find the largest number $A$ for which the following statement is true.

No matter how we choose seven real numbers between 1 and $A$, there will always be two of them for which the ratio $h$ satisfies $\frac{1}{2} \leqq h \leqq 2$.
Consider the closed intervals $[1 ; 2], [2 ; 4], [4 ; 8], [8 ; 16], [16 ; 32], [32 ; 64]$. If $A \leq 64$ and is a natural number greater than $1$, then no matter how we choose seven real numbers between 1 and $A$, there will always be at least one of the above six intervals that contains at least two of these seven numbers. The ratio of these two numbers will then satisfy the condition. Thus, the sought number is at least 64.

If $A > 64 = 2^6$, then $\sqrt[6]{A} > 2$. Therefore, the ratio of any two integer powers of $\sqrt[6]{A}$ is greater than 2 or less than $1/2$. Since the 0th, 1st, 2nd, ..., 6th powers of $\sqrt[6]{A}$ lie between 1 and $A$, and the ratio of any two of these seven numbers does not satisfy the condition, the sought number $A$ cannot be greater than 64.

Thus, the sought number is 64.

Remark. We get the same result if we interpret the word "between" in the text to exclude equality. The first part of the proof can be repeated word for word, and in the case of $A > 64$, we need to start from a number $B$ between 64 and $A$—that is, $A > B > 64$. There exists a number $C$ such that $C > 1$, and $C \cdot B$ is still less than $A$. If we now multiply $C$ by the first seven non-negative integer powers of $\sqrt[6]{B}$, which is greater than 2, each of the resulting seven numbers will be greater than 1 and less than $A$, and the ratio of any two of them will not satisfy the stated condition.