14. The smallest two $n$ that satisfy the condition are $n=2,6$. Based on this analysis, we prove the following strengthened proposition: there exist infinitely many positive even numbers $n_{1}<n_{2}<\cdots$ such that for any $k \in \mathbf{N}^{*}$, we have $n_{k} \mid\left(2^{n_{k}}+2\right)$, and $\left(n_{k}-1\right) \mid\left(2^{n_{k}}+1\right)$.

We use mathematical induction to prove this proposition. Let $n_{1}=2$, and for any $k \in \mathbf{N}^{*}$, let $n_{k+1}=2^{n_{k}}+2$. We claim that the sequence defined in this way satisfies the proposition.

When $k=1$, it is clearly true. Assume that $n_{k} \mid\left(2^{n_{k}}+2\right)$ and $\left(n_{k}-1\right) \mid\left(2^{n_{k}}+1\right)$. Then $n_{k+1}=2^{n_{k}}+2$ is even, and $2^{n_{k}}+2=n_{k} \cdot q$, where $q$ is a positive odd number. Furthermore, let $2^{n_{k}}+1=\left(n_{k}-1\right) \cdot p$, where $p$ is a positive odd number. We have
$$\begin{aligned} 2^{n_{k+1}}+2 & =2\left(2^{n_{k+1}-1}+1\right)=2\left(2^{\left(n_{k}-1\right) p}+1\right)=2\left(2^{n_{k}-1}+1\right) M \\ & =\left(2^{n_{k}}+2\right) M=n_{k+1} \cdot M \end{aligned}$$

where $M \in \mathbf{N}^{*}$ (by factorization). Thus, $n_{k+1} \mid\left(2^{n_{k+1}}+2\right)$.
On the other hand, we also have
$$2^{n_{k+1}}+1=2^{n_{k} \cdot q}+1=\left(2^{n_{k}}+1\right) \cdot N$$

where $N$ is a positive integer. Thus, $\left(n_{k+1}-1\right) \mid\left(2^{n_{k+1}}+1\right)$. Therefore, the strengthened proposition holds.