Definition We will denote by $ a\rightarrow b$ the fact that there exists a positive integer $ k$ and a finite sequence $ n_{1},\cdots, n_{k}$ of positive integers such that $ n_{1}= a$, $ n_{k}= b$, and $ n_{i}n_{i+1}$ is divisible by $ n_{i}+n_{i+1}$ for each $ i$ ($ 1\le i\le k$). Observation 1 If we reverse the sequence of $ k$ positive integers such that $ n_{1}= a$, $ n_{k}= b$, and $ n_{i}n_{i+1}$ is divisible by $ n_{i}+n_{i+1}$ for each $ i$ ($ 1\le i\le k$), we get a sequence of $ k$ positive integers such that $ n_{1}= b$, $ n_{k}= a$, and $ n_{i}n_{i+1}$ is divisible by $ n_{i}+n_{i+1}$ for each $ i$ ($ 1\le i\le k$). Therefore, $ a\rightarrow b\Leftrightarrow b\rightarrow a$. Also, note that if $ (x+y)|xy$ then $ x\rightarrow y$ and $ y\rightarrow x$. Lemma 1 $ a\rightarrow a(a-1)$. Proof $ a+a(a-1) = a^{2}|a^{2}(a-1) = (a)(a(a-1))\Rightarrow (a+a(a-1))|(a)(a(a-1))$ so $ a\rightarrow a(a-1)$ and the lemma is proved. Lemma 2 If $ a$ is odd, then $ a(a-1)\rightarrow a(a+1)$. Proof $ a(a-1)+a(a+1) = 2a^{2}|a^{2}(a-1)(a+1) = (a(a-1))(a(a+1))$ because $ a-1$ is even as $ a$ is odd. So, $ (a(a-1)+a(a+1))|(a(a-1))(a(a+1))$ and the result of lemma 2 follows. Lemma 3 $ a\rightarrow b, b\rightarrow c\Rightarrow a\rightarrow c$. Proof We can just take the finite sequence $ n_{1},\cdots, n_{k}$ of positive integers such that $ n_{1}= a$, $ n_{k}= b$, and $ n_{i}n_{i+1}$ is divisible by $ n_{i}+n_{i+1}$ for each $ i$ ($ 1\le i\le k$) and the a finite sequence $ m_{1},\cdots, m_{l}$ of positive integers such that $ m_{1}= b$, $ m_{l}= c$, and $ m_{i}m_{i+1}$ is divisible by $ m_{i}+m_{i+1}$ for each $ i$ ($ 1\le i\le l$), and the sequence $ a = n_{1}, n_{2},...,n_{k}= b = m_{1}, m_{2},...,m_{l}= c$ will satisfy the fact that the sum of very two consecutive terms in the sequence will divide their product, so $ a\rightarrow c$ and lemma 3is proved. Lemma 4 For any positive integer $ a\geq 3$, $ 3\rightarrow a$. Proof We will prove the result using mathematical induction. Base Step If $ a = 3$ then $ a\rightarrow 3\Rightarrow 3\rightarrow 3$ which is true. Induction step Assume, for $ i = 3,4,...,k$ we have $ 3\rightarrow i$. Let us prove $ 3\rightarrow k+1$. If $ k+1$ is even, $ k$ is odd, so by lemmas 1 and 2, have $ k+1\rightarrow k(k+1)\rightarrow k(k-1)\rightarrow k$ ($ k(k-1)\rightarrow k$ because of observation 1 and $ k\rightarrow k(k-1)$ by lemma 1). Hence, by lemma 3 $ k+1\rightarrow k$, so $ k\rightarrow k+1$. But $ 3\rightarrow k, k\rightarrow k+1$ so by lemma 3 $ 3\rightarrow k+1$ and the result is true for $ a = k+1$. If $ k+1$ is odd, by lemmas 1 and 2 have $ k+1\rightarrow k(k+1)\rightarrow (k+2)(k+1)\rightarrow k+2$ so by lemma 3, $ k+1\rightarrow k+2$ so $ k+2\rightarrow k+1$. Because $ k+1$ is odd, $ k+2$ is even, let $ k+2 = 2l$, then $ k = 2l-2$. Now, $ (2l+2l(l-1))|(2l*2l(l-1))\Rightarrow 2l(l-1)\rightarrow 2l = k+2$. Also, $ (2l(l-1)+2(l-1)(l-2)) = 4(l-1)^{2}|(2l(l-1))(2(l-1)(l-2))\Rightarrow 2(l-1)(l-2)\rightarrow 2l(l-1)$. Finally, $ (2(l-1)+2(l-2)(l-1)) = 2(l-1)^{2}|(2(l-1)*2(l-2)(l-1))\Rightarrow 2(l-1)\rightarrow 2(l-2)(l-1)$. Because $ k+2 = 2l$, then $ k = 2l-2 = 2(l-1)$. So, $ k = 2l(l-1)\rightarrow 2(l-2)(l-1)\rightarrow 2l(l-1)\rightarrow 2l = k+2\rightarrow k+1\Rightarrow k\rightarrow k+1$, but because $ 3\rightarrow k$ by lemma 3 it follows that $ 3\rightarrow k+1$ and the result is true for $ a = k+1$. Hence, in both cases we have proved the result for $ a = k+1$ and by mathematical induction it follows that $ 3\rightarrow a$ for any positive integer $ a\geq 3$ and lemma 4 is proved. Now, for any two positive integers $ a,b$ greater than $ 2$, by lemma 4, $ 3\rightarrow a\Leftrightarrow a\rightarrow 3$ and $ 3\rightarrow b$. So, $ a\rightarrow 3$ and $ 3\rightarrow b$ so by lemma 3, $ a\rightarrow b$ and the result of the problem follows.

First, we can see that if there is a finite sequence satisfying the problem requirements such that $n_{1} = a$ and $n_{k} = b$, then we can make a sequence that satisfies the problem requirements such that $m_{1} = b$ and $m_{k} = a$ with the simple mapping of $m_{j} = n_{k-j+1}$. Also, note that if we have a term X in our sequence and D is a divisor of X then (D-1)X may be the next term in the sequence because $\frac{(D-1)X^2}{DX}$ is an integer. Thus if X is a term in our sequence, then we can make X! a term in our sequence as well. If we want to construct a sequence that takes a to b, we can assume that a is less than b. Now, we can construct a sequence that takes a to a! and we now want to construct a sequence that takes a! to b. Let us consider the case in which b is odd. We can see that b+1 is now even and thus divisible by 2. Also, let us note that we can take a! to (2a)! by simply taking a! to 2*a! and noting that this is divisible by 2a which lets us take it to (2a)!. If $a < \frac{b+1}{2} \rightarrow 2a < b+1$. We can now take a! to $(2^k*a)!$ such that $\frac{b+1}{2} \le 2^k*a < b+1$. We can now note that 2 and $\frac{b+1}{2}$ both divide $(2^k*a)!$ thus we can take $(2^k*a)!$ to $b*(2^k*a)!$ and take this to b!. However, we can take b! to b because we can take b to b!. Now let us consider the case in which b is even. We can take b to b(b-1) to (b-2)(b-1) to b-1 which is odd and thus we can construct a path from a to b-1 to b. Thus we can always construct a path from a to b for any pair of integers (a, b).