To solve the problem, we analyze the divisors of a composite integer $n$ and determine for which $n$ the divisibility condition holds.

Let $n$ be a composite integer with the positive divisors $d_1, d_2, \ldots, d_k$ such that $1 = d_1 < d_2 < \cdots < d_k = n$. We must check that for every $1 \leq i \leq k - 2$, $d_i$ divides $d_{i+1} + d_{i+2}$.

First, consider the case where $n = p^m$ for some prime $p$ and integer $m \geq 2$. The divisors of $n$ are $1, p, p^2, \ldots, p^m$.

For each $i$, the divisibility condition is:
$$d_i \mid d_{i+1} + d_{i+2}.$$
Substituting the divisors gives:
$$p^{i-1} \mid p^i + p^{i+1}.$$
Simplifying, we have:
$$p^{i-1} \mid p^i (1 + p),$$
which holds true because $p^{i-1}$ clearly divides $p^i$.

Therefore, if $n = p^m$ for some prime $p$ and integer $m \geq 2$, the condition is satisfied.

Now, assume $n$ has at least two distinct prime factors, say $n = p^a q^b$ for distinct primes $p$ and $q$. The divisors include $1, p, q, pq, p^2, q^2,\ldots$.

Consider $n = 6 = 2 \cdot 3$ as a small example. The divisors are $1, 2, 3, 6$, and for $i = 1$, $d_1 = 1$ should divide $d_2 + d_3 = 2 + 3 = 5$, which it does. For $i = 2$, $d_2 = 2$ should divide $d_3 + d_4 = 3 + 6 = 9$, which is not divisible by 2.

Hence, having multiple distinct prime factors can violate the divisibility condition, verifying that only numbers of the form $n = p^m$ satisfy the given property.

The solution is that $n$ must be of the form:
$$\boxed{n = p^m \text{ for some prime } p \text{ and integer } m \geq 2.}$$