IMO Shortlist

Answer: $n=p^{r}$ is a prime power for some $r \geqslant 2$.

Solution 1. It is easy to see that such an $n=p^{r}$ with $r \geqslant 2$ satisfies the condition as $d_{i}=p^{i-1}$ with $1 \geqslant i \geqslant k=r+1$ and clearly
$$p^{i-1} \mid p^{i}+p^{i+1}$$
Now, let us suppose that there is a positive integer $n$ that satisfies the divisibility condition of the problem and that has two different prime divisors $p$ and $q$. Without loss of generality, we assume $p<q$ and that they are the two smallest prime divisors of $n$. Then there is a positive integer $j$ such that
$$d_{1}=1,\ d_{2}=p,\ \ldots,\ d_{j}=p^{j-1},\ d_{j+1}=p^{j},\ d_{j+2}=q$$
and it follows that
$$d_{k-j-1}=\frac{n}{q},\ d_{k-j}=\frac{n}{p^{j}},\ d_{k-j+1}=\frac{n}{p^{j-1}},\ \ldots,\ d_{k-1}=\frac{n}{p},\ d_{k}=n$$
Thus
$$\begin{equation*} d_{k-j-1}=\frac{n}{q} \left\lvert\, d_{k-j}+d_{k-j+1}=\frac{n}{p^{j}}+\frac{n}{p^{j-1}}=\frac{n}{p^{j}}(p+1)\right. \tag{1} \end{equation*}$$
This gives $p^{j} \mid q(p+1)$, which is a contradiction since $\operatorname{gcd}(p, p+1)=1$ and $p \neq q$.

Solution 2. Since $d_{i} d_{k+1-i}=n$, we have the equivalence:
$$d_{k-i-1}\left|d_{k-i}+d_{k-i+1} \Longleftrightarrow \frac{n}{d_{i+2}}\right| \frac{n}{d_{i+1}}+\frac{n}{d_{i}}$$
We multiply both sides by $d_{i} d_{i+1} d_{i+2}$ and cancel the $n$'s to get
$$d_{i} d_{i+1} \mid d_{i} d_{i+2}+d_{i+1} d_{i+2}$$
Hence,
$$\begin{equation*} d_{i} \mid d_{i+1} d_{i+2} \tag{2} \end{equation*}$$
Moreover, by the condition of the problem,
$$d_{i} \mid d_{i+1}\left(d_{i+1}+d_{i+2}\right)=d_{i+1}^{2}+d_{i+1} d_{i+2}$$
Combining this with (2) we get that $d_{i} \mid d_{i+1}^{2}$ for all $1 \leqslant i \leqslant k-2$.
Let $d_{2}=p$ be the smallest prime divisor of $n$. By induction on $i$ we prove that $p \mid d_{i}$ for all $2 \leqslant i \leqslant k-1$. The base case $d_{2}=p$ is obvious. Let us suppose that $p \mid d_{j}$ for some $2 \leqslant j \leqslant k-2$. Then we have that
$$p\left|d_{j}\right| d_{j+1}^{2} \Longrightarrow p \mid d_{j+1}$$
as $p$ is prime, which completes the induction. This implies that $n$ has to be a prime power, as otherwise there would be another prime $q$ that divides $n$ and we would get that $p \mid q$ which is obviously false.
We finally check that the powers of $p$ satisfy the condition in the statement of the problem as in Solution 1.

Solution 3. We start by proving the following claim:
Claim. $d_{i} \mid d_{i+1}$ for every $1 \leqslant i \leqslant k-1$.
Proof. We prove the Claim by induction on $i$; it is trivial for $i=1$ because $d_{1}=1$. Suppose that $2 \leqslant i \leqslant k-1$ and the Claim is true for $i-1$, i.e. $d_{i-1} \mid d_{i}$. By the induction hypothesis and the problem condition, $d_{i-1} \mid d_{i}$ and $d_{i-1} \mid d_{i}+d_{i+1}$, so $d_{i-1} \mid d_{i+1}$.
Now consider the divisors $d_{k-i}=\frac{n}{d_{i+1}}, d_{k-i+1}=\frac{n}{d_{i}}, d_{k-i+2}=\frac{n}{d_{i-1}}$. By the problem condition,
$$\frac{d_{k-i+1}+d_{k-i+2}}{d_{k-i}}=\frac{\frac{n}{d_{i}}+\frac{n}{d_{i-1}}}{\frac{n}{d_{i+1}}}=\frac{d_{i+1}}{d_{i}}+\frac{d_{i+1}}{d_{i-1}}$$
is an integer. We conclude that $\frac{d_{i+1}}{d_{i}}$ is an integer, so $d_{i} \mid d_{i+1}$.
By the Claim, $n$ cannot have two different prime divisors because the smallest one would divide the other one. Hence, $n$ must be a power of a prime, and powers of primes satisfy the condition of the problem as we saw in Solution 1.

Solution 4. We present here a more technical way of finishing Solution 1 after obtaining (1). We let $v_{p}(m)$ denote the $p$-adic valuation of $m$. Notice that $v_{p}(n / q)=v_{p}(n)$ as $\operatorname{gcd}(p, q)=1$ and that
$$v_{p}\left(\frac{n}{p^{j}}(p+1)\right)=v_{p}(n)-j$$
as $\operatorname{gcd}(p, p+1)=1$. But (1) implies
$$v_{p}(n)=v_{p}(n / q) \leqslant v_{p}\left(\frac{n}{p^{j}}(p+1)\right)=v_{p}(n)-j$$
which is a contradiction. Thus $n$ has only one prime divisor as desired.