Proposed by Mykhailo Shtandenko and me. You can find our original solution in the shortlist; here's a much better one found by Srinivas Arun. Let $a_m = \sqrt{d_1 + d_2 + \ldots d_m}$. Then $a_m + a_{m-1} \mid a_m^2 - a_{m-1}^2 = d_m$, so $a_m+a_{m-1}$ is a divisor of $n$. Since all such numbers are distinct, they're a permutation of $d_1, d_2, \ldots, d_k$. Since $a_m + a_{m-1} \le d_m$, we must in fact have $a_m + a_{m-1} = d_m$ for all $m$, so the divisors of $n$ are consecutive odd integers. This leaves $n=1$ and $n=3$, which work.

The answer is $n=1$ and $n=3$ which work. Note that $d_1 = 1$, since no larger perfect squares differ by $1$. The following claim will then overkill the problem: Claim: $d_m = 2m-1$ for every $m$. Proof. Proceeding by induction, suppose $d_1 = 1$, $d_2 = 3$, \dots, $d_m = 2m-1$. Thus \[ d_1 + d_2 + \dots + d_m = m^2. \]Let $d_1 + \dots + d_m + d_{m+1} = (m+u)^2$ for some $u$, so that \[ d_{m+1} = (m+u)^2 - m^2 = u(2m+u). \][asy][asy] size(12cm); draw( (-1,0)--(17,0), grey, Arrows); draw( (0,0)..(0.5,0.5)..(1,0), blue, EndArrow(TeXHead), Margins ); label("$d_1=1$", (0.5,0.5), dir(90), blue); draw( (1,0)..(2.5,0.5)..(4,0), blue, EndArrow(TeXHead), Margins ); label("$d_2=3$", (2.5,0.5), dir(90), blue); draw( (4,0)..(6.5,0.5)..(9,0), blue, EndArrow(TeXHead), Margins ); label("$d_m=2m-1$", (6.5,0.5), dir(90), blue); draw( (9,0)..(12.5,0.5)..(16,0), red, EndArrow(TeXHead), Margins ); label("$d_{m+1}=u(2m+u)$", (12.5,0.5), dir(90), red); dot("$0$", (0,0), dir(-90)); dot("$1$", (1,0), dir(-90)); dot("$4$", (4,0), dir(-90)); dot("$m^2$", (9,0), dir(-90)); dot("$(m+u)^2$", (16,0), dir(-90)); [/asy][/asy] Assume for contradiction that $u > 1$. Then $2m+u$ is a divisor of $d_{m+1}$, hence a divisor of $n$, which does not appear among $\{d_1, d_2, \dots, d_{m+1}\}$. However, any two perfect squares greater than $(m+u)^2$ differ by strictly more than $2m+u$, so $2m+u$ cannot appear among $\{d_{m+2}, \dots, d_k\}$ either, contradiction. $\blacksquare$

Let $d_1+\ldots+d_i=a_i^2$ for all $i$. Then, note that $d_i=a_i^2-a_{i-1}^2$ for all $i \geq 1$, and so $1,a_1+a_2,a_2+a_3,\ldots,a_{k-1}+a_k$ are all divisors of $n$, as are $a_1,a_2^2-a_1^2,\ldots,a_k^2-a_{k-1}^2$. Since in both groups all numbers are distinct ($a_i<a_j$ if $i<j$ and so $1<a_1+a_2<a_2+a_3<\ldots<a_{k-1}+a_k$), and there are exactly $k$ numbers in each group, they are both permutations of all divisors of $n$, hence they have an equal product: $(a_1+a_2)(a_2+a_3)\ldots(a_{k-1}+a_k)=a_1(a_2^2-a_1^2)\ldots(a_k^2-a_{k-1}^2)$, and so $a_1(a_2-a_1)\ldots(a_k-a_{k-1})=1$, which implies that $a_1=1$ and $a_{i+1}-a_i=1$ for all $i$. Thus, $a_i=i$ for all $i$, which in turn umplies that $d_i=a_i^2-a_{i-1}^2=2i-1$ for all $i$. Hence, all divisors of $n$ are $1,3,\ldots, n=2k-1$. However, if $n>1$ ($n=1$ clearly works), we would have that $n-2 \mid n$, hence $n=3$, which works too. To sum up, all solutions are $1$ and $3$.

Let $a_i^2=d_1+d_2+\ldots+d_i$. Then, since $a_{j+1}^2-a_j^2\geq2a_j+1\geq2a_i+1$ for $J\geq i$, we must have $d_{j+1}\geq2a_i+1$. We will prove that $d_i=2i-1$ for $1\leq i\leq k$ by induction on $i$. Base Case: $i=1$ Then, since $1$ divides $n$, we must have $d_j=1$ for some $j$. If $j\geq2$, then $d_j\geq2a_{j-1}+1\geq3$, which is impossible. Therefore, $j=1$, so $d_1=1$. Inductive Step: We have $d_1+\ldots+d_i=(d_1+\ldots+d_{i-1})+d_i=(i-1)^2+d_i$ by the inductive hypothesis. Since $a_i^2-(i-1)^2=d_i$, we must have $a_i+(i-1)\mid a_i^2-(i-1)^2\mid n$, so $d_j=a_i+i-1$. Since $a_i>i-1$, we get $d_j<2a_i<2a_i+1$, which implies $j-1<i$, so $j\leq i$. Therefore, $a_i+(i-1)\leq2i-3$ or $a_i+(i-1)=a_i^2-(i-1)^2$. The first case becomes $a_i\leq i-2$, which is absurd. Therefore, $a_i-(i-1)=1$, so $a_i=i$, which implies $d_i=i^2-(i-1)^2=2i-1$. Therefore, $d_k=2k-1=n$, so $d_{k-1}=2k-3\mid n$. This implies $k=1$ or $2k-3\mid 2k-1$, so $2k-3\mid 2$. Therefore, $k=1$ or $k=2$, which gives the two solutions $\boxed{n=1}$ and $\boxed{n=3}$, which both work.

Pretty much the same induction, but still will post. The answer is $n=1,3$ Let $s_i^2=d_1+d_2+...d_i$. The main claim is that $s_i=i$ and $d_i=2i-1$ for all $i$. Then it's easy to finish: note that $n-2 | n$. We prove it by induction on $i$, the base case is trivial. Before we begin, obviously $s_i \geq i$ and $d_i=s_i^2-s_{i-1}^2 \geq s_i+s_{i-1} \geq 2i-1$. Suppose now we are proving it for $i+1$. Hence, $d_{i+1}=s_{i+1}^2-s_i^2=s_{i+1}^2-i^2$. The key is to note that now $s_{i+1}+i\geq 2i+1$ is also a divisor, and the only possibility for it is $d_{i+1}$ because it is bigger than $d_i$ but not bigger than $d_{i+1}$. Now plugging back gives $s_{i+1}=i+1$ so we are done.

Yay, another my problem in the shortlist, btw one of the 4 Ukrainian problems in this Shortlist. Also, two problems were sent by me to IMO 2021 in my first year after graduation from high school and one appeared in the IMO, one in the shortlist . I came up with the statement and Fedir helped me to prove it, also thanks to Oleksii Masalitin, who encouraged me to create some non-geo problems, and I came up with this. I think, I'd rather prefer to see this problem at the IMO instead of G7.

The only possible values of $n$ are $n=1$ and $n=3$. It is easy to verify that they indeed work. Observe that $d_1=1$, since there are no two positive integer squares differing by $1$. Also let $a_i^2 = d_1+d_2 +\ldots+d_i$ for every positive integer $1 \le i \le k$. It is not hard to see that sequence $a_l$ is increasing. Claim: $d_2 =3$. Proof: . Note that $d_2=(a_2-1)(a_2+1)$. Suppose that $d_2$ is composite number. Observe that $a_2+1 \mid d_2$ and $a_2+1 \neq 1$. This means $a_2+1$ is divisor of $n$ that has not appeared yet. Note that: $$ a_2^2 <a_2^2+a_2+1 < (a_2+1)^2 $$As the distances between squares get bigger, we can conclude that $a_2+1$ cannot be divisor of $n$. Therefore $d_2$ is prime number, meaning $a_2-1=1 \implies d_2=3$ as desired. We will prove by mathematical induction that $d_l =2l-1$ with base case already being done. Suppose that $d_i =2i-1$ for all $i \le k$, then we need to prove that $d_{k+1}=2k+1$. Note that: $$ a_k =1+3+\ldots+2k-1 = k^2$$. The main idea is as follows. Note that: $$ d_{k+1} =(a_{k+1}-k)(a_{k+1}+k) $$Firstly, obserbe that $a_{k+1} \ge k+1$. Moreover that $a_{k+1}+k \ge 2k+1$ meaning that it is some divisor $d_j$ of $n$ with $j>k+1$. In the case $j=k+1$, we have $a_{k+1}=k+1 \implies d_{k+1}=2k+1$ as needed. Secondly, note that: $$ d_j = (a_{j+1}-a_j)(a_{j+1}+a_j) $$The crucial observation is that $$ a_{k+1}+k = d_j \ge a_{j+1}+a_j > a_{k+1}+a_{k} \implies k>a_k =k$$where we have used the fact that $a_l$ is an increasing sequence. Contradiction. This means that all divisors of some number $N$ are $1;3;5; \ldots; N-2,N$. But $N-2 \mid N \implies N-2 \mid 2 \implies N=3$ or $N=1$.

Define the sequence $a_i$ such that $d_1=a_1^2, d_2=a_2^2-a_1^2, d_3=a_3^2-a_2^2, \dots, d_k=a_k^2-a_{k-1} ^2$ with $1\leq a_1<a_2<\dots <a_k$, note that if $n$ is even then there will be $g$ such that $a_g^2-a_{g-1}^2\equiv 2(mod 4)$, absurd. Note that if $d_1\neq 1$ then there is $j$ such that $d_j=1=a_j^2-a_{j-1}^2=(a_j-a_{j-1})(a_j+a_{j -1})\Rightarrow a_j=1, a_{j-1}=0$, absurd, so $d_1=1$. So $d_2=a_2^2-1=(a_2-1)(a_2+1)$, so that if $p$ is an odd prime that divides $a_2+1$ then there is $d$ such that $a_d ^2+p=a_{d+1}^2\Rightarrow p=2a_d+1\leq a_2+1$, where the last inequality comes from the divisibility limiting property, but as $a_3>a_2$ we have $a_d= a_1=1\Rightarrow p=3\Rightarrow 3^k=a_2+1$, however, since $q$ is an odd prime that divides $a_2-1$ in an analogous way, we have the existence of $r$ such that $a_r ^2+q=a_{r+1}^2\Rightarrow q=2a_r+1\leq a_2-1\Rightarrow a_r=1$ so that $3|a_2-1$(absurd by $mod3$) or $a_2 =2\Rightarrow d_2=3$. Now we have $d_3+4=a_3^2\Rightarrow d_3=(a_3-2)(a_3+2)$, if $s$ is an odd prime that divides $a_3+2$, then there exists $i$ such that $ a_i^2+s=a_{i+1}^2\Rightarrow s=2a_i+1\leq a_3+2\Rightarrow i\in \{1,2\}\Rightarrow s=3$ or $s=7$ , but acting in an analogous way when taking a prime divisor of $a_3-2$ we arrive at $d_3=7^a\cdot 3^b$, where we have two cases: 1. $a\geq 1 \Rightarrow 7|n \Rightarrow$ exists $k$ such that $7=(a_k-a_{k-1})(a_k+a_{k-1})\Rightarrow a_k=4, a_{k -1}=3$ but as $2=a_2<a_3<\dots <a_k$ we will have $k=4 \Rightarrow d_3=3^2-2^2=5$, absurd. 2. $a=0\Rightarrow d_3=3^b\Rightarrow 4+3^b=a_3^2\Rightarrow (a_3-2)(a_3+2)=3^b\Rightarrow x=3\Rightarrow$ absurd. So there is no $d_3 \Rightarrow n=1$ or $n=3$.

The answer is $n=1$ and $n=3$, which obviously work. Let the divisors of $n$ in strictly increasing order be $a_1,a_2, \dots, a_k$ Let $x_i:=\sqrt{d_1+d_2+ \cdots d_i}$ with $x_0=0$. We have that $x_i+x_{i-1}|x_i^2-x_{i-1}^2|d_i|n$. Thus $x_1,x_2+x_1, \dots , x_k+x_{k-1}$ are all divisors of $n$ and since this is a strictly increasing sequence we have that $x_i+x_{i-1}=a_i \implies a_i=d_i$ and so $\{d_i \}$ is a strictly increasing sequence. This means that $x_1=d_1=1$. The following claim obviously finishes the problem. Claim: $d_i=2i-1$. Proof: We use induction with the base case being proven. Thus assume $d_m=2m-1$ for all $1 \le m \le i$. Then we must have that $d_{i+1}=(i+l)^2-i^2 = l(2i+l)$. If $l>1$, then $2i+l$ is not in $d_1,d_2, \dots, d_{i+1}$, and so $2i+l$ must be later on than $d_{i+1}$ a contradiction to the fact that $\{d_i \}$ is a strictly increasing sequence so we are done.

The answers are $n = 1$ and $n = 3$, which work with the permutations $(1)$ and $(1, 3)$. Now we will show that no other $n$ work. Since $1$ is always a divisor of $n$, there exists a $1\leq i\leq k$ such that $d_i = 1$. Assume for the sake of contradiction that $i\neq 1$. Then, $d_1 + d_2 + \cdots + d_{i-1} = a_0^2$ for some positive integer $a_0$ and $a_0^2 + 1 = d_1 + d_2 + \cdots + d_{i-1} + d_i = b_0^2$ for some positive integer $b_0$, so $(b_0-a_0)(b_0+a_0) = b_0^2 -a_0^2 = 1$, i.e. $(b_0-a_0, b_0+a_0) = (1, 1)$ or $(-1, -1)$, giving $(a_0, b_0) = (0, 1); (0, -1)$ respectively, a contradiction (since $a_0$ and $b_0$ are positive, as the $d_i$ are as well). Thus, $d_1 = 1$. If $k=1$, then this means that the permutation is $(1)$, so $n$ must be $1$, else $n > 1$ and so $n$ would be included in the permutation. Now assume $k > 1$. Claim: For $1\leq i\leq k$, $d_i = 2i-1$. Proof: Strong induct on $i$. We have already established the base case $i = 1$ from our work above. Now, assume the result holds for $i = 1, 2, \cdots, j$ and consider $d_{j+1}$. We know that $$a^2 = d_{j+1} + d_j + \cdots + d_1 = d_{j+1} + (2j-1) + (2j-3) + \cdots + 1 = d_{j+1} + j^2$$for some positive integer $a$. Then, $d_{j+1} = a^2 - j^2 = (a-j)(a+j)$, so $a > j$. Assume for the sake of contradiction that $a \geq j+2$. Then, $a - j \geq 2$, so $a + j \mid d_{j+1}$ and $a+j < d_{j+1}$, so it must be included somewhere else in this list of divisors, say $a_p = a+j$ for some positive integer $p$. As we said, $p \neq j+1$. If $p\leq j$, then $2j < a + j = a_p = 2p-1 \leq 2j-1$, a contradiction. If $p > j+1$ (if such a $p$ exists in our permutation), then $a_1 + a_2 + \cdots + a_{p-1}$ and $a_1 + a_2 + \cdots + a_p$ are perfect squares, say $c^2$ and $d^2$ for positive integers $c, d$ with $d > c$. These are both greater than or equal to $a_1 + a_2 + \cdots + a_{j-1} = a^2$, so $d > c > a$. Their difference is $a+j = a_p$, so $a+j = d^2 - c^2 = (d-c)(d+c) \geq d+c > 2a\iff j > a$, but $a > j$, a contradiction. Thus, $a = j+1$ so $d_{j+1} = 2j + 1 = 2(j+1) - 1$, as desired. Thus, $(d_1, d_2, \cdots, d_k) = (1, 3, 5, \cdots, 2k-1)$. Since $n$ is in this list and greater than all of the other numbers in the list (the other numbers are its proper divisors), it follows that $n = 2k-1$. However, since we assumed $k > 1$, $d_{k-1} = 2k-3$ is also in this list, so it must be a divisor of $n$, ergo $2k-3\mid 2k-1\iff 2k-3 \mid 2$, so $2k-3 \in \{1, 2\}$, giving $k = 2$ and $k = 2.5$ respectively (the latter is absurd), and $k = 2$ gives $n = 2k-1 = 3$ which indeed works, so we have shown that if $k = 1$, the only solution is $n = 1$, and if $k > 1$, the only solution is $n = 3$, which both work, so we're done.

Let $s_r^2=d_1+d_2+\dots+d_r$ We claim that $d_i=2i-1.$ We will proceed with induction. First, we'll show that $d_1=1.$ Suppose that $d_i=1$, then $s_{r+1}^2-s_{r}^2=1$ which implies $s_r=0$, so $i=1.$ Obviously $1$ appears somewhere in $d$ so $d_1=1.$ $~$ Now, suppose $d_1=1,d_2=3,\dots,d_r=2r-1.$ Then, $r^2+d_{r+1}=(r+s)^2$ then $d_{r+1}=s^2+2rs.$ It remains to show that $d_{r+1}=s+2r.$ Note that $s+2r\mid n$ so it must appear in the $d$ sequence. Obviously it doesn't appear before $d_{r+1}$ so suppose $d_{r+k}=s+2r$ then $s_{r+k}^2-s_{r+k-1}^2=s+2r.$ However, $s_{r+k}^2-s_{r+k-1}\ge s_{r+k}+s_{r+k-1}> 2r+2s$ and so $d_{r+1}=s+2r$ $~$ Thus, $s^2+2rs=s+2r$ implies $s=1$, so the claim holds. Note that this implies the only answers are $n=1,3$ which work.

The answer is $n=1,3$, which clearly works. The idea is to consider gaps between perfect squares and prove the key claim that $\boxed{d_m=2m-1}$ for every $m$. This is by induction, with the base case of $d_1=1$ being clear as we otherwise have two positive perfect squares with difference $1$. Now suppose that $d_i=2i-1$ for $i=1,\ldots,m$, hence $$d_1+\cdots+d_m=m^2.$$If $d_1+\cdots+d_{m+1}=(m+a)^2$ for some $x>0$, then $d_{m+1}=(m+a)^2-m^2=a(2m+a)$. If $a>1$, then $2m+a \mid n$, yet $2m+a \not \in \{d_1,\ldots,d_{m+1}\}$. Hence there must exist some $x>y\geq m+a$ such that $x^2-y^2=2m+a$. However, we have $$x^2-y^2\geq (y+1)^2-y^2=2y+1\geq 2m+2a+1$$for all such $(x,y)$: contradiction. This forces $a=1$, so $d_{m+1}=2m+3$. This means that the divisors of $n$ are $1,3,\ldots,2k+1$ for some $k$, hence $2k-1 \mid 2k+1 \implies 2k-1 \mid 2$, which is only possible for $k=0,1 \implies n=1,3$ as desired. $\blacksquare$

We claim that the answer is $n\in \left \{ 1;3\right \}$ which clearly works. Let $a_0=0$ and $a_i=\sqrt{a_{i-1}^2+d_i}$ for $i>0.$ For every $m=1,2,...,n$ we have $b_m=a_m+a_{m-1}|a_m^2-a_{m-1}^2=d_m|n,$ so $b_1,b_2,...,b_k$ form a permutation of $d_1,d_2,...,d_k$ and by $b_m\leq d_m$ we in fact obtain $b_m=d_m\implies a_m=a_{m-1}+1.$ By trivial induction $d_m=2m-1\implies n-2=d_{k-1}|n\implies n\in \left \{ 1;3\right \}.$

The answer is $n=1,3$ let $d_1+d_2+ \cdots +d_i = m_i^2$. First, we see $d_1=m_1=1$ since the only two squares with difference $1$ are $0^2$ and $1^2$. Notice that $(m_i+m_{i-1})(m_i-m_{i-1})=m_i^2-m_{i-1}^2 = d_i$ which divides $n$. So, $m_i+m_{i-1} > 1$ is always a divisor of $n$. But since $m_1, \ldots, m_k$ is increasing, $m_{i-1}+m_i \not = m_{j-1}+m_j$ if $i \not = j$. So, $\{ 1,m_1+m_2,m_2+m_3, \ldots, m_{k-1}+m_k \}$ is the set of all divisors of $n$ since there exactly $k$ elements. But also recall that $\{ 1, m_2^2-m_1^2, m_3^2-m_2^2, \ldots, m_k^2-m_{k-1}^2 \}$ is also the set of all divisors of $n$. Notice that the only value of $i$ where \[ m_{i-1}+m_i = m_k^2-m_{k-1}^2 \geq 2m_{k-1}+1\]is $i=k$ since all other choices are too small. Inductively, we go down looking at the largest remaining sum $m_{\ell-1}+m_{\ell}$ and conclude that it must be equal to the largest remaining difference of square $m_{\ell}^2-m_{\ell-1}^2$ in order to obey the size constraints. So $m_{i+1}^2-m_i^2=m_{i+1}+m_i$ for all $i$. Thus, we have $m_{i+1}-m_i=1$ for all $i$. Since $m_1=1$ it follows that $m_t=t$ for all $t$. So, $d_t = 2t-1$ for all $t$. This clearly only generates a set of divisors if $k=1,2$ yielding $n=1,3$ as desired.

We claim $n=1$ and $n=3$ are the only solutions. Assume there is such a sequence. Let the sequence $a_i$ be defined as the square root of the sum of the first $i$ terms. We have that \[a_i+a_{i+1}\mid{}a_{i+1}^2-a_i^2=d_i\]However, because no two $d_i$ are equal and no two $a_i+a_{i+1}$ are equal, $a_i+a_{i+1}$ is equal to $d_i$ itself, giving us that $a_{i+1}-a_i$ is always $1$. Therefore our only solutions are $n=1$ and $n=3$, and we are done.

Let $f(n)$ be the number of prime divisors of $n$, counting multiplicity. In particular, $f(1)=0$. The answer is $n=\boxed{1,3}$, the permutations being $(1)$ and $(1,3)$. Claim. $n$ is odd. Proof. $x^2+2=y^2$ has no solutions. Claim. $d_i$ takes $\left(\dfrac{d_i-1}{2}\right)^2$ to $\left(\dfrac{d_i+1}{2}\right)^2$. Proof. We strong induct on $f(d_i)$. The base case, $d_i=1$, is trivial. Suppose that this statement is true whenever $f(d_i)\le m$. Now consider a $d_i$ such that $f(d_i)=m+1$. So \[x^2+d_i=y^2\rightarrow d_i=(y-x)(y+x).\]But note that if one of these factors isn't $1$, then say $y-x=s$ and $y+x=t$ where $st=d_i$. But $f(t)\le m$, so this is going to overlap with it, contradiction. So $y-x=1$ and $y+x=d_i$, and we are done. Now note that this means $d_i$ have to cover all odd positive integers less than some cutoff, and that is obviously only possible with $n=1$ and $n=3$(by Bertrand's, say). QED.

Let $a_m=\sqrt{d_1+\ldots+d_m}$ for $1 \le m \le k$. Then \[ a_m+a_{m-1} \mid a_m^2-a_{m-1}^2 = d_m \mid n. \]Thus, each $a_m+a_{m-1}$ is $d_{f(m)}$ for some function $f: [k]\backslash\{1\} \rightarrow [k]$. In particular, $f(m) \le m$ for each $2 \le m \le k$. Since $f$ is strictly increasing and $f(2)=2$, it is clear by induction that $f(m)=m$. Now, since $a_m+a_{m-1}=d_m=a_m^2-a_{m-1}^2$ for all $2 \le m \le k$, $a_m-a_{m-1}=1$. Thus, $a_{\ell}=\ell$ for all $1 \le \ell \le k$, and $d_{\ell}=2\ell-1$ for these values of $\ell$ too. Finally, assume $n>5$ and consider $d_{k-1}$ and $d_{k-2}$. Clearly $\gcd(d_{k-1}, d_{k-2})=1$, so \[d_{k-1}d_{k-2}=\text{lcm}(d_{k-1}, d_{k-2}) \mid n, \]which is not possible due to size reasons. Testing $1 \le n \le 5$, we find that $\boxed{n=1}$ and $\boxed{n=3}$ are the only solutions.

Since $1$ is a divisor of $n$, clearly, $d_1 = 1$. Let $a^2 = d_1 + d_2$. Then, $d_2 = (a-1)(a+1)$. If $a \neq 2$, there must be some $j > 2$ such that $d_j = a-1$; however, the difference between any two distinct perfect squares at least $a^2$ will be at least $2a+1$, so there is nowhere to put $a-1$. This is a contradiction, so $a = 2$. Hence, $d_1 = 1$ and $d_2 = 3$. It follows that $n$ is odd. We claim that the perfect squares $d_1 + d_2 + \cdots + d_i$ must all be consecutive. In particular, we will show that for $i = 1,2, \dots k-1$, the perfect squares $d_1 + d_2 + \dots + d_i$ and $d_1 + d_2 + \dots + d_{i+1}$ are consecutive. We will use strong induction on $i$, with a base case of $i$ such that $d_i$ is prime, and an inductive step of $i$ such that $d_i$ has one more not necessarily distinct prime divisor than the step before. (okay this makes no sense lol.) For the base case, clearly, if $d_i$ is prime, by difference of squares, it must be the difference of two consecutive squares: $\left( \frac{d_i-1}{2} \right)^2$ and $\left( \frac{d_i+1}{2} \right)^2$. For the inductive step, we consider some $d_i$, assuming all its divisors are the difference between two consecutive squares in the ordering. We will show that $d_i$ also must be the difference between two consecutive squares. By difference of squares, any pair $(a,b)$ such that $ab = d_i$ corresponds with the perfect squares $\left( \frac{a-b}{2} \right)^2$ and $\left( \frac{a+b}{2} \right)^2$. However, if $b > 1$, we have a contradiction, since by our inductive assumption, the perfect squares $\left( \frac{a-1}{2} \right)^2$ and $\left( \frac{a+1}{2} \right)^2$ were consecutive. Thus, $b=1$, completing the induction. (I will offer a 900 dollar cash prize to anybody who understood what the above even means, including myself.) There is clearly no positive integer whose divisors are $1,3,5, \dots, 2k+1$ for any $k>1$, so the only solution is $n=1,3$.

The answer is $n=1$ and $n=3$. It is clear these work. Let $a_i = \sqrt{d_1+d_2+\dots + d_i}$. We induct to show $d_i$ has to be $2i-1$. The base case is obvious. For the inductive step, assume $d_i = 2i-1$ for $i<k$. We want to show $d_k = 2k-1$. Let $a_k = m$. Then $d_k = m^2-(k-1)^2 = (m-k+1)(m+k-1)$. Obviously $m+k-1$ is also a factor of $n$. If $m\neq k$, then $m+k-1 < 2m -1 = (m+1)^2 - m^2$, so it must have been used already. But it is at least $2k-1$, so $d_k$ must be exactly $2k-1$. Then $n$ is a number which has its factors as a prefix of the odd numbers. It is obvious this only occurs for $n=1,3$.

Let $x_i^2 = d_1 + d_2 +\cdots +d_i$. First, note that $1 = d1$, as if it were at some other position $i$, $x_i^2 + 1 = x_{i+1}^2$, impossible. Next, we have clearly $d_i = x_i^2 - x_{i-1}^2 = (x_i + x_{i-1})(x_i - x_{i-1})$ for $2\ leq i \leq k$. Then since $x_i + x_{i-1}$ and $x_i - x_{i-1}$ are divisors of $d_i$, they must also be divisors of $n$. Additionally, since $x_k > x_{k-1} >\cdots x_1$, the quantity $x_i + x+{i-1}$ is distinct across $i$. (However we cant say the same about the other term, so we ignore it). Since each of the $n-1$ divisors $n$ larger than $1$ are divisible by $n-1$ distinct other divisors of $n$ larger than $1$, these two sets are congruent. Also, since $d_k > d_{k-1} >\cdots d_2 > 1$ and $x_k + x+{k-1} > x_{k-1} + x_{k-2} >\cdots x_2 + x_1$, we must have $d_i = x_i + x_{i-1}$ for all $i$, so, $x_i - x_{i-1} = 1$. This clearly means that $d_i = 2i-1$, and so $n$ 's divisors are all consecutive odd integers. However, this implies $n$ is odd and has $n-2$ as a divisor, and this is only possible if $n = 3$, and we also include the simple case of $n = 1$.

Let $n$ be such a positive integer. For every $i=1,2,\ldots,k$, we let $a_i=\sqrt{d_1+d_2,\ldots,d_i}$, so that $(a_i)$ is an increasing sequence of positive integers. For convenience we also let $a_0=d_0=0$. Note that, for every positive integer $0\le i\le k-1$, we have \[d_{i+1}=a_{i+1}^2-a_i^2=(a_{i+1}-a_i)(a_{i+1}+a_i)\quad (*).\]Since $d_{i+1}$ divides $n$ for every such $i$, it must be the case that $a_{i+1}+a_i$ also divides $n$ for every such $i$. Thus, the $k$ numbers \[1=a_1<a_1+a_2<a_2+a_3<\ldots<a_{k-1}+a_k=n\]are exactly the $k$ positive divisors of $n$, in increasing order. In particular we have $a_1=1$, so $d_1=1$. Note that this also implies that the sequences $(d_i)$ and $(a_{i-1}+a_i)$ are the same up to permutation. Now take the product of equation $(*)$ for every $i=0,1,\ldots,i-1$ to get that \[\prod_{i=1}^k d_i=\prod_{i=0}^{k-1}(a_{i+1}-a_{i})\cdot \prod_{i=0}^{k-1}(a_i+a_{i+1}),\]so $\prod_{i=0}^{k-1}(a_{i+1}-a_{i})=1$ and it follows that $a_{i+1}-a_i=1$ for all $i$ since every factor is a positive integer. Starting from $a_1=1$ we get by induction that $a_i=i$ for all $i$, and thus \[a_{k-2}+a_{k-1}=2k-3\mid n=a_{k-1}+a_k=2k-1,\]or in other words $k\in \{1,2\}$, which respectively give $n=1$ and $n=3$. Conversely, $n=1$ and $n=3$ are the only positive integers that satisfy the problem condition. $\blacksquare$

Same as other solutions. Sketch: Establish $n$ is odd, then induct on $d_i$to prove $d_i=2i-1$ which gives $n=1,3$ as the only solutions.

The answer is $n=1$ and $n=3$ only. Obviously $d_1=1$, and as $1+d_2 = i^2$ for some prime $d_2$, we must have $d_2 = 3$. Now, consider building up a sequence $(d_1, d_2, \dots)$ constructively such that $d_1 < d_2 < \dots$ and for $d_i$ and $d_j$ in the sequence, $d_id_j$ must also be in the sequence. Claim. $d_i = 2i-1$ for each $i$. Proof. We proceed inductively to show this is the only possibility. Note that $d_1+d_2+\cdots+d_k = k^2$ by the inductive hypothesis, and because $3$ is in the sequence, $3d_k = 6k-3 \geq 2k-1$ must also be in the sequence, which implies that $d_{k+1} \leq 6k-3$, or $d_{k+1} = 2k+1$ is fixed as it must be odd. $\blacksquare$ However, $\{2i+1\}_1^n$ is not the complete set of divisors of any positive integer for any $n \geq 2$, which implies the result.

cool bruh Claim 1: $2 \nmid n$: Say, for the sake of contradiction, $2 \mid n \implies d_i = 2$, which implies that: $$d_1 + ... + d_{i - 1} + 2 = k^2$$Since $d_1 + ... + d_{i - 1}$ is also a perfect square (say $m^2$), we have $k^2 - m^2 = 2$, which is clearly not possible. Claim 2: $d_1 = 1$: SFTSOC, $d_1 \neq 1$. We know $d_1$ must be a perfect square (say $a^2$) $\implies a^2 \mid n \implies a | n$. Say $a = d_i$. $$\implies a^2 + ... + d_{i - 1} + a = x^2$$Since $a^2 + ... + d_{i - 1}$ is also a perfect square (say $y^2$), we have $x^2 - y^2 = a = (x - y)(x + y) \geq (a + 1 + a) = 2a + 1$. This is clearly not possible, which means $a^2$ and $a$ are indistinct, $a = 1$. So, $d_1 = 1$. Claim 3: $d_2 = 3$ Say $1 + d_2 = a^2 \implies (a - 1)(a + 1) = d_2$. So, we have $a - 1, a + 1 \mid n$. Either $d_2 = a + 1$, or $a + 1$ is some other $d_i$ where $i > 2$. Take the latter case. We have: $$1 + d_2 + ... + d_{i - 1} + (a + 1) = k^2$$Take $1 + d_2 + ... + d_{i - 1} = l^2 \implies k^2 - l^2 = a + 1 = (k - l)(k + l) \geq 2a + 1$, which is not possible. Therefore, $d_2 = a + 1 \implies a + 2 = a^2 \implies a = 2 \implies d_2 = 3$. Claim 4: $d_i = 2i - 1$ We use induction. Say $\{d_1, ... d_i\} = \{1, ..., 2i - 1\}$. We have: $$1 + ... + 2i - 1 + d_{i + 1} = i^2 + d_{i + 1} = k^2 \implies (k - i)(k + i) = d_{i + 1}$$This means that $k + i$ is a divisor of $n$. Note that $k + i \geq 2i + 1$, which means it's after $d_{i}$. Hence, $d_{i + 1} = k + i$, or $k + i$ is after $d_{i + 1}$. Say $k + i = d_m$. We take a look at the latter case first: $$1 + ... + 2i - 1 + d_{i + 1} + ... + d_{m - 1} + d_m = z^2$$Say $1 + ... + 2i - 1 + d_{i + 1} + ... + d_{m - 1} = x^2 \implies k + i = z^2 - x^2 = (z - x)(z + x)$. Since $x \geq k \geq i + 1$, $z \geq k$, we have $k + i \geq k + i + 1$, which is clearly false. Therefore, we must have $k + i = d_{i + 1}$, which means $k + i = k^2 - i^2 \implies k - i = 1 \implies k = i + 1$, so $d_{i + 1} = 2i + 1 = 2(i + 1) - 1$, which completes. Since $d_k = n$, $d_{k - 1} = n - 2$, we have $n - 2 | n \implies n - 2 | 2 \implies n = 1, 3$. It clearly works, so we are done.

Let $a_i^2 = d_1 + d_2 + \dots + d_i$. Obviously, we have $d_1 = 1$, so $a_1 = 1$. Now note that $a_i + a_{i-1} \mid a_i^2 - a_{i-1}^2 = d_i$, hence $a_i + a_{i-1}$ is a divisor of $n$. Since $1 < a_1 + a_2 < a_2 + a_3 < \dots < a_{k-1} + a_k$ are all divisors of $n$, we have $\{1, a_1 + a_2, \dots a_{k-1} + a_k\} = \{d_1, d_2, \dots, d_k\} = \{1, a_2^2 - a_1^2, \dots a_k^2 - a_{k-1}^2\}$. Therefore $(a_1 + a_2)(a_2 + a_3) \cdots (a_{k-1} + a_k) = (a_2^2 - a_1^2)(a_3^2 - a_2^2) \cdots (a_k^2 - a_{k-1}^2)$, which means $a_i - a_{i-1} = 1$ for all $i$. Hence $a_i = i$ for all $i$, thus $d_i = 2i - 1$ for all $i$, meaning $n = d_k = 2k-1$. But if $k \ge 2$, then we have $2k-3 \mid 2k-1$, so $2k - 3 = 1$, which means $n = 2k - 1 = 3$. Otherwise, we get $n = 1$. Therefore the only answers are $n = 1, 3$. $\blacksquare$

We show by strong induction that $d_i=2i-1.$ For the base case if $d_1=m^2\ne 1$ then $m$ cannot be added later since the gaps between squares are too big. For the inductive step assume $d_i=2i-1$ for $1\le i\le m.$ Then the sum is at $m^2.$ Suppose the next square is $(m+k)^2$ for $k>1.$ Then $k(2m+k)$ is a factor of $n,$ so $2m+k$ is a different factor of $n.$ By the inductive hypothesis all currently added factors are $<2m$ so $2m+k$ is not yet added but the smallest gap between squares possible now is $(m+k+1)^2-(m+k)^2=2m+2k+1>2m+k,$ impossible, thus completing the induction. Now if $n$ has $k$ factors with $k>3$ then they must be $1,3,\dots,2k-1,$ and Bertrand's guarantees there is some prime dividing $n$ with $p \ge k.$ Then $3p\ge 3k>2k-1$ is a factor of $n,$ contradiction. Checking $k=1,2,3$ we see that $k=1,2$ give $n=1,3$ which work.

We show that the only solutions are $n=1$ and $n=3$, which clearly work. Note that there are no solutions in positive integers to $x^2-y^2=1$. Hence, $d_1=1$ is forced. Similarly, there are no solutions to $x^2-y^2=2$, so $n$ must be odd. Claim: $d_m=2m-1$ for each $1\leq m\leq k$. Proof. We proceed using induction. The base case $m=1$ is done. We induct on $m$. Suppose $d_{m+1}\ne 2m+1$. We have: \[d_1+d_2+\cdots+d_m+d_{m+1}=m^2+d_{m+1}=(m+u)^2\Longrightarrow d_{m+1}=u(2m+u)\]Clearly $u>1$. Note that $u$ must be odd or else $2\mid n$, which is not possible. Now, $2m+u$ is a divisor of $n$ and does come in the sums. Consider $x^2-y^2=2m+u=(x-y)(x+y)$. This means $y\leq m+\frac{(u+1)}2$. However, we know $y^2\geq (m+u)^2$, so $y\geq m+u$, which means $2u\leq u+1$ an evident contradiction. The induction is complete. $\blacksquare$ Finally, note that $n-2\mid n$, which means $n-2\mid 2$ which is impossible unless $n=1, 3, 4$. Now, $n$ is odd as well, so only $n=1,3$ work, as claimed. $\blacksquare$

We will show that $d_i$ must equal $2i - 1$, by induction on $i$ with our base cases being true. Let $d_1 + d_2 + \dots + d_i = s_i^2$. So then we have $d_{i+1} = s_{i+1}^2 - s_i^2 \implies s_{i+1} + s_i \mid n \implies \{s_i\} = \{d_i\}$. Now applying the induction hypothesis, we get that $d_{i+1} = 2(i+1) - 1$ as desired. Now notice that the only two numbers so that $1, 3, 5, \dots, 2k + 1$ cover all divisors of some number $2k + 1$ are obviously only $1$ and $3$, so we are done. This is since for sufficiently large $n$(> $4$), $2k - 1$ is greater than half of $2k + 1$, so it cannot be a divisor.

Very nice problem; seems about average 2/5 difficulty because there are so many things you can try. Always try stupid things first! The only answers are $n=1$ and $n=3$, which both work. Observe that $n$ must be odd, as no two perfect squares differ by $2$ (so we cannot have $d=2$ for any $d \mid n$.) Let $S_k^2 = d_1+d_2+\cdots+d_k$ for each $k$. Claim. We have $d_m = 2m-1$ for all $m \geq 1$. Proof. We will induct on $m$. Clearly $d_1 = 1$, as no two perfect squares differ by $1$. Now, notice suppose that $d_i = 2i-1$ for all $i \leq k-1$ and $d_k \neq 2k-1$. Then $S_k> k$; in particular, \[d_1+d_2+\cdots+d_k = (k+\ell)^2 \geq (k+2)^2\]by the parity condition. Note that we have \[d_k = (k+\ell)^2 - (k-1)^2 = (\ell + 1)(2k+\ell - 1)\]so in particular $2k+\ell - 1 \mid n$, so there is some $j$ such that $d_j = 2k+\ell - 1$. But observe that $\ell \geq 2$, so $d_j \not \in \{d_1, d_2, \dots, d_{k-1}\}$. Furthermore, for any $j > k$, we have \[d_j = S_j^2-S_{j-1}^2 \geq 2S_{j-1} + 1 \geq 2 S_k + 1 = 2k+2\ell-1 > 2k+\ell-1.\]Since we clearly have $d_k > d_j$ too, $d_j$ cannot be in the sequence, which yields a contradiction. Remark: In a sense, this problem is very rigid, unexpectedly; I spent way too much time trying to find global properties relating to the $\{d_k\}$ instead of attempting to directly characterize them.

I will induct that it must be $1$ or $3$ or divisible by $1, 3, \dots $, suppose it is divisible by $1, 3, \dots, 2n-1$ for some $n$, than if that is not all the divisors of a number the next divisor must be $2n+1$, this is because we have that if the next square is $(n+u)^2$ we get that the next divisor is $u(n+u)$, however as $n+u<u(n+u)$ if $u>1$ this is a contradiction.

The answer is $n=1$ and $n=3$ only, which clearly work. Clearly, no two squares differ by $2$, so $n$ must be odd. The main claim is the following. Claim: Every divisor $d$ of $n$ must increase the square root by $1$, that is, go from a prefix sum of $(\frac{d-1}{2})^2$ to a new sum of $(\frac{d+1}{2})^2$. We induct on the size of $d$. For $d=1$, this is clearly true. Then, suppose $d$ takes it from $a^2$ to $b^2$ with $b^2-a^2=(b+a)(b-a)=d$. If $d$ is prime, then we must have $b+a=p$ and $b-a=1$, so the claim is true. Then, consider otherwise. We must have $d_1,d_2$ with $d_1>d_2$ and $d_1d_2=d$ such that $b+a=d_1$, $b-a=d_2$. We wish to show that $d_1=d$ and $d_2=1$. This means that $a=\frac{d_1-d_2}{2}$ and $b=\frac{d_1+d_2}{2}$, in other words, we go from $(\frac{d_1-d_2}{2})^2$ to $(\frac{d_1+d_2}{2})^2$. We claim that we must have $d_1=d$. If not, then $d_1$ is a divisor of $n$ smaller than $d$, and thus the inductive hypothesis applies and we must go from $(\frac{d_1-1}{2})^2$ to $(\frac{d_1+1}{2})^2$ when $d_1$ is added. However, if $d_1<d$ and $d_2>1$, then this contradicts the fact that we also have to go from $(\frac{d_1-d_2}{2})^2$ to $(\frac{d_1+d_2}{2})^2$, since this jump entirely contains the former. Thus we have shown the claim. Thus, the factors of $n$ must be consecutive odd integers. For $n\geq 5$, $n-2$ does not divide $n$. Thus we must have $n=1$ or $n=3$.