, 2022

The answer is $n = 1, 3$.
For $i = 1, 2, \dots, k$ let $d_1 + \dots + d_i = s_i^2$, and define $s_0 = 0$ as well. Obviously
$0 = s_0 < s_1 < s_2 < \dots < s_k$, so $s_i \ge i$ and
$$d_i = s_i^2 - s_{i-1}^2 = (s_i + s_{i-1})(s_i - s_{i-1}) \ge s_i + s_{i-1} \ge 2i - 1. \quad (1)$$
The number 1 is one of the divisors, hence we must have $d_1 = 1$.
Consider $d_2$ and $s_2 \ge 2$. By definition, $d_2 = s_2^2 - 1 = (s_2 - 1)(s_2 + 1)$, so the numbers $s_2 - 1$ and $s_2 + 1$ are divisors of $n$. In particular, there is some index $j$ such that $d_j = s_2 + 1$.
Notice that
$$s_2 + s_1 = s_2 + 1 = d_j \ge s_j + s_{j-1}; \qquad (2)$$
since the sequence $s_0 < s_1 < \dots < s_k$ increases, the index $j$ cannot be greater than 2. Hence, the divisors $s_2 - 1$ and $s_2 + 1$ are listed among $d_1$ and $d_2$. Thus $s_2 - 1 = d_1 = 1$ and $s_2 + 1 = d_2$; therefore, $s_2 = 2$ and $d_2 = 3$. By repeating the above procedure, we can prove that $d_i = 2i - 1$ and $s_i = i$ for $i = 1, 2, \dots, k$.
Suppose we already have $d_j = 2j - 1$ and $s_j = j$ for all $j \ge i$. Consider $d_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2 = (s_{i+1} - i)(s_{i+1} + i)$, the number $s_{i+1} + i$ is a divisor of $n$, so there is some index $j$ such that $d_j = s_{i+1} + i$.
Similarly to (2), by (1) we have
$$s_{i+1} + s_i = s_{i+1} + i = d_j \ge s_j + s_{j-1} \qquad (3)$$
since the sequence $s_0 < s_1 < \dots < s_k$ increases, (3) forces $j \le i + 1$. On the other hand, $d_j = s_{i+1} + i > 2i > d_i > d_{i-1} > \dots > d_1$, so $j \le i + 1$ is not possible. The only possibility is $j = i + 1$.

$$\begin{aligned} s_{i+1} + i &= d_{i+1} = s_{i+1}^2 - s_i^2 = s_{i+1}^2 - i^2; \\ s_{i+1}^2 - s_{i+1} &= i(i+1). \end{aligned}$$
By solving the equation, we get $s_{i+1} = i + 1$ and $d_{i+1} = 2i + 1$.
As a result, the divisors of the number $n$ must be $1, 3, \dots, n-2, n$ and $d_k = 2k-1 = n$, which is an odd number. Observe that $d_{k-1} = n-2$ is a divisor of $n$, so we have $n \le 4$. Thus the only possible $n$ is 1 or 3. Can easily check both satisfies the desired property.