Question 1-1 When any two of the positive divisors of a positive integer $n$ are added together, it is found that these sums are all different, and they are arranged from smallest to largest as $4$, $6$, $8$, $10$, $12$, $14$, $16$, $18$, $20$, $24$, $46$, $48$, $50$, $54$, $60$. Find the value of $n$.
Problem $1-1$ has a flavor of reverse thinking.
From the known conditions, we can derive some useful information, such as the fact that "the sum of divisors has 15" implies that $n$ has six positive divisors, and according to the sum of each divisor being even, we know that $n$ must be an odd number, etc. In fact, to more directly determine the value of $n$, we can set the smallest prime factor of $n$ to be $p$. Then its smallest two positive divisors must be $1, p$, and the largest two positive divisors must be $n, \frac{n}{p}$.
Thus, $1+p=4, n+\frac{n}{p}=60$.
Consequently, $n=45$.