Determine all composite integers $n>1$ that satisfy the following property: if $d_1<d_2<\cdots <d_k$ are all the positive divisors of $n$, then $d_i$ divides $d_{i+1}+d_{i+2}$ for every $1\le i\le k-2$.

Step 1 (Prime powers work): If $n=p^{\alpha }$, divisors are $1,p,p^2,\dots ,p^{\alpha }$. Check: $p^i|p^{i+1}+p^{i+2}=p^{i+1}(1+p)$. Since $\gcd (p,1+p)=1$, $p^i|p^{i+1}$ which is true. So all prime powers work.

Step 2 (Two distinct prime factors fail): Suppose $n$ has two distinct prime factors $p<q$. The divisors include $1,p,q,\dots$. Consider $d_1=1,d_2=p,d_3=q$ (or the next divisor). We need $1|p+q$ (trivially true). We need $p|q+d_4$ where $d_4$ is the fourth-smallest divisor.

If $n=pq$ (no higher powers), divisors are $1,p,q,pq$. Need $p|q+pq=q(1+p)$. Since $\gcd (p,q)=1$: $p|(1+p)$, so $p|1$, giving $p=1$. Contradiction. So $n=pq$ fails.

Step 3 (General two-prime case): For $n=p^a{q}^b$, the first few divisors are $1,p,\dots ,q,\dots$ The divisibility condition on early triples forces contradictions similar to Step 2.

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