Choose an odd $n$ with at least $2(k+1)$ positive factors. $n=a_1b_1=...=a_{k+1}b_{k+1}$. Write $a_ib_i=x_i^2-y_i^2$. $x_1^2-y_1^2=....=x_{k+1}^2-y_{k+1}^2$. $x_i^2-x_1^2=y_i^2-y_1^2$. So, $(x_i-x_1)(x_i+x_1)=(y_i-y_1)(y_i+y_1)=n_i$ The $n_i$ we chose satisfy the condition.

61plus wrote: Let $D(n)=\{ a-b\mid ab=n, a>b>0, a,b\in\mathbb{N} \}$. Prove that for any $k$ there exists pairwise distinct positive integers $n_1, n_2,\ldots ,n_k$, such that $|D(n_1)\cap D(n_2)\cap\cdots\cap D(n_k)|\geq 2$. For the integer $n>1$ , define $D(n)=\{ a-b\mid ab=n, a>b>0, a,b\in\mathbb{N} \}$. Prove that: For any integer $k>1$ , there exists pairwise distinct positive integers $n_1, n_2,\ldots ,n_k$ and $n_1>1, n_2>1,\cdots ,n_k>1$, such that $|D(n_1)\cap D(n_2)\cap\cdots\cap D(n_k)|\geq 2$.

Over-complicated Solution For any two distinct integers $x,y$ define $$S(x,y)=\{n|\{x,y\}\subset D(n)\}$$Obviously the question is equivalent to showing $|S(x,y)|$ is unbounded. CLAIM. $$a_x=|S(4x-1,x-4)|$$is unbounded. Proof. Let $S_x=S(4x-1,x-4)$ Firstly, Notice that $4x\in S_x$. Now notice that if some integers $a,b$ satisfies $$a(a-(x-4))=b(b-(4x-1))\quad (1)$$Then this common value belongs to $S_x$. Now $(1)$ is equivalent to $$(a-x)(a+4)=(b-4x)(b+1)\quad (2)$$Now fix a pair of $(m,n)$. Then any pair of $a,b$ satisfying the following system will satisfy $(2)$ \begin{align*} m(a-x)&=n(b+1)\\ n(a+4)&=m(b-4x)\\ \end{align*}Solving the above system we get $$(m^2-n^2)b=4xm^2+(4+x)mn+n^2=(4m+n)(xm+n)$$Now since $(m+n,m-n)=1$ it suffices to find $m,n$ such that \begin{align*} m+n&|(4m+n)(xm+n)\\ m-n&|(4m+n)(xm+n) \end{align*}This is equivalent to \begin{align*} m+n|3(x-1)\\ m-n|5(x+1)\\ \end{align*}Now by Chinese Remainder Theorem we can find $x$ such that $x$ is odd and $(3(x-1),5(x+1))=1$ and that $p_i|x+1$ for some $i=1,2,...,m$. Then $5(x+1)$ will have a lot of prime factors. For each of them we can obtain a different pair of $(m,n)$, hence a different pair of $a,b$. This completes the proof.