Hint

Nice! Odd $n$ does not work, since $n=2k+1$ does not have two divisors that are $k$ apart. Hence $n=2m$ must be even. The difference $k=m$ is covered by $k=(2m)-(m)$. The difference $k=m-1$ is covered by $k=(m)-(1)$. The difference $k=m-2$ is covered by $k=(m)-2$. The difference $k=m-3$ is either covered by (a) $k=(m)-3$ or (b) $k=(m-1)-2$ or (c) $k=(m-2)-1$. (b) and (c) only allow solutions with very small $n$. (a) needs further consideration.

Suppose $n\ge2$ has $d$ distinct divisors. Then $n\ge2^{d/2}$. The $d$ divisors yield $\binom{d}{2}$ pairs and at most $\binom{d}{2}$ distinct differences. The problem statement requires $\binom{d}{2}\ge n/2\ge 2^{d/2}/2$. This yields $d\le 200$, and brings the problem down to checking a finite (though impractical) range.

I will finish a).We then have that $n=6m$.Now,consider $k=2m+1$.That can only be covered with $(3m)-(m-1)$,so $m-1$ divides $6m$ so we need to check $m=1,2,3,4,7$.

If there exists a positive integer $p \leq \lfloor n/6 \rfloor$ such that $p\nmid n$, then we have $\lfloor n/2 \rfloor > \lfloor n/6 \rfloor$, and taking $k=\lfloor n/2 \rfloor-p \geq 2$ and two positive divisors $d,d+k$ of $n$, we need $d+(\lfloor n/2 \rfloor-p)$ to divide $n$. But $d+(\lfloor n/2 \rfloor-p) \geq d+\lfloor n/2 \rfloor - \lfloor n/6 \rfloor > d+(n/2 -1) - n/6 \geq n/3$, so $d+(\lfloor n/2 \rfloor-p) \in \{n/2,n\}$, the only possible divisors of $n$ larger than $n/3$. However, $d+(\lfloor n/2 \rfloor-p) = n/2$ yields $d=p$, absurd (since $d\mid n$ but $p\nmid n$), while $d+(\lfloor n/2 \rfloor-p) = n$ yields $d>n/2$, thus $d = n$ (since $d\mid n$), forcing $p=\lfloor n/2 \rfloor > \lfloor n/6 \rfloor$, again absurd. Therefore all positive integers not larger than $\lfloor n/6 \rfloor$ must divide $n$. Denote $u = \lfloor n/6 \rfloor$. Since $\gcd(u,u-1)=1$, it follows $u(u-1) \mid n$, so $u(u-1) \leq n = 6(n/6) < 6(u+1)$, forcing $u \leq 7$. For $u\geq 4$ we need $\operatorname{lcm}[1,2,3,4] = 12\mid n$, and we can see that $n=24$ satisfies, and moreover is an acceptable value. For $n=36$ we get $u=6$, but $\operatorname{lcm}[1,2,3,4,5,6] = 60\nmid n$. And for $n\geq 48$ we have $u\geq 8$, not acceptable. Thus the answer is $\boxed{n=24}$. We may in fact quite easily exhibit the full set $\{1,2,4,6,8,12,18,24\}$ of such positive integers $n$ (for $n = 1$ the condition is vacuously fulfilled). The related question of which are the positive integers satisfying the above property for all $1\leq k \leq n-1$ can also easily be answered; the full set is $\{1,2,4,6\}$.

manuel153 wrote: Suppose $n\ge2$ has $d$ distinct divisors. Then $n\ge2^{d/2}$. Unfortunately that claim is false. A simple counterexample uses $n=2^a3^b$. Then $d=(a+1)(b+1)$. Your claim is that $2^a3^b \geq 2^{(a+1)(b+1)/2}$, thus $3^b \geq 2^{(a+1)(b+1)/2-a}$. But $\dfrac {(a+1)(b+1)}{2} - a = \dfrac {(a+1)(b-1)}{2}+1$, and for $b>1$ that can easily be larger than $2b$ if we take $a$ large enough. However, then $3^b < 2^{2b} = 4^b$. The problem requires a subtler argument.

Hmm it might be easier to write if we use small cases to show $2,3 \mid n$ first. Then, it's easy to see that all integers less than or equal to $\frac{n}{2}-\frac{n}{3} = \frac{n}{6}$ must divide $n$, so $\frac{n}{6} \cdot \left(\frac{n}{6}-1\right) = n \cdot \frac{n-6}{36} \le n \leftrightarrow n \le 30$, and it is easy to check $n = 30$ is unacceptable. Is there a way to generalize the case for $k \le \frac{n}{a}$?

why $3 \mid n$ first ?

First, note that 24 works. Then it follows that n must be divisible by 6 for n/2 -2 and n/2 -3 to work (which are greater than n/3 cuz n is greater than 24). Also note that all numbers between n/2 -1, n/2 -2, ... n/3 must have n/2 as one subtractand the other is a divisor smaller than n/2. So our condition reduces to: Lcm(1, 2, ... n/6) $\mid$ n. For n/6 = 5, We can test by hand and see that the statement does not hold. For n/6 >5 it suffices to use the inequality Lcm(1, 2, ... n/6) > (n/6)(n/6 -1)(n/6 -2)/2 > n.

Bernard pastupalute

Note 1. Two greatest positive divisors of $n$ and differ from it do not exceeding $\frac{n}{2}$ and $\frac{n}{3}$, respectively. Obviously. Note 2. If $n \geq 6$ then $n$ is even. Else we can't get any odd number between 1 and $\frac{n}{2} \geq 3$, including 3. $\square$ Note 3. Checking the small cases $n \leq 12$ gives us $n \in \{1, 2, 4, 6, 8, 12\}$. Now let assume $n = 2k$ where $k \geq 7$. Then b/c of $0 < k - 3 < k$ there are two divisors $d_i$ and $d_j$ of $n$ with difference $k - 3$. Case 1. $k - 1 | n$. Then $$k - 1 = (k - 1, 2k) = (k - 1, 2)$$therefore $k \leq 3$, contradiction. Case 2. $k - 2 | n$. Then $$k - 2 = (k - 2, 2k) = (k - 2, 4)$$which follows $k \leq 6$, contradiction. It remains to check the last case $3 | n$. Let $n = 6l$ ($l \geq 3$). Using Note 1 we can try to find two divisors which difference is exactly $2l + 1$. It can be only $3l$ and $l - 1$. So $$l - 1 = (l - 1, 6l) = (l - 1, 6)$$which implies $l \leq 7$. The answer: 24.