For $n=3,k=2$ Let $n>3$ Then $k-1|n,n-k|n$ but $max(k-1,n-k) \geq \frac{n-1}{2}>\frac{n}{3}$ so $n=2max(k-1,n-k)$ or $n=k-1$ $n=2k-2 \to k-2|2k-2 \to k-2|2 \to k=3,n=4$ or $k=4,n=6$ $n=2n-2k \to n=2k$ contradiction because $0$ should be divisor $k=n+1 $ Let $d>1,d\leq \frac{n}{2}$ is divisor of $n$ then $n+1-d|n \to n+1-d \leq \frac{n}{2} \to n+2\leq 2d \to$ contradiction. So $n$ is prime Answer $n=3,4,6$ or prime

Answer: n is prime, or n=4, or n=6 Proof: For every prime number n, k=n+1 works. For n=4, k=3 works. For n=6, k=4 works. Assume that n is not a prime. Let p be the smallest prime divisor of n. k-1|n, implies 2<=k<=n+1 If k=n+1, then n+1-p|n, hence n+1-p|p-1 but n-p+1>=2p-p+1=p+1>p-1. a contradiction. Hence, 2<=k<=n n-k|n Hence, n-k|k, implying n/2<=k<=n-1 The largest two divisors of n are n, and n/p=m k-m|n. Hence k-m>=p or k=m+1 k-p|n. Hence k-p<=m Implying: k=m+1 or k=m+p If k=m+1, then, if p>2 n-k|n n-m-1|n But p(n-m-1)=pn-n-p=n(p-1)-p>=n+n-p>n. A contradiction. p=2 n=2m k=m+1 k-2=m-1|2m m-1|2 m=2 or 3 n=4 or 6. If k=m+p k-1|n m+p-1|n m+p-1>m, hence, m+p-1=n. k=n+1, which contradicts k<=n-1. Done.

Isn't $n=1$ solution as well (one can just take $k=2$)?