Proof. Let $a_1, a_2, \dots, a_{k+1}$ be $k+1$ distinct positive integers, where each of them is smaller than the product of other $k$ numbers. Write $N = a_1 a_2 \cdots a_{k+1}$. For each $i = 1, 2, \dots, k+1$, let $x_i = \frac{1}{2} \left( \frac{N}{a_i} + a_i \right)$, $y_i = \frac{1}{2} \left( \frac{N}{a_i} - a_i \right)$, then $x_i^2 - y_i^2 = N$.
Since $a_i a_j < N$ and $\frac{N}{a_i} > a_i$, $(x_i, y_i)(1 \le i \le k+1)$ are $k+1$ positive integer solutions of equation $x^2 - y^2 = N$. Without loss of generality, suppose $x_{k+1} = \min\{x_1, x_2, \dots, x_{k+1}\}$. For each $i \in \{1, 2, \dots, k\}$, since $x_i^2 - y_i^2 = x_{k+1}^2 - y_{k+1}^2$, we have
$$(x_i + x_{k+1})(x_i - x_{k+1}) = x_i^2 - x_{k+1}^2 = y_i^2 - y_{k+1}^2 = (y_i + y_{k+1})(y_i - y_{k+1}).$$
Let $n_i = (x_i + x_{k+1})(x_i - x_{k+1}) = (y_i + y_{k+1})(y_i - y_{k+1})$, then
$$2x_{k+1} = (x_i + x_{k+1}) - (x_i - x_{k+1}) \in D(n_i),$$
$$2y_{k+1} = (y_i + y_{k+1}) - (y_i - y_{k+1}) \in D(n_i).$$
So $x_{k+1} > y_{k+1}$, $2x_{k+1}$ and $2y_{k+1}$ are two different members of $D(n_1) \cap D(n_2) \cap \dots \cap D(n_k)$. $\square$