For a subset $S$ of $\{1, \ldots, 2014\}$, let $\mathcal{F}_{S}$ be the set of all sets $T$ such that $S \subseteq T \subseteq \{1, \ldots, 2014\}$. It can be checked that the sets $\mathcal{F}_{S}$ satisfy the conditions 1 and 2. We claim that the $\mathcal{F}_{S}$ are the only sets of subsets of $\{1, \ldots, 2014\}$ satisfying the conditions 1 and 2. (Thus, the answer is the number of subsets $S$ of $\{1, \ldots, 2014\}$, which is $2^{2014}$.) Suppose that $\mathcal{F}$ satisfies the conditions 1 and 2, and let $S$ be the intersection of all the sets of $\mathcal{F}$. We claim that $\mathcal{F}=\mathcal{F}_{S}$. First, by definition of $S$, all elements $T \in \mathcal{F}$ are supersets of $S$, so $\mathcal{F} \subseteq \mathcal{F}_{S}$. On the other hand, by iterating condition 1, it follows that $S$ is an element of $\mathcal{F}$, so by condition 2 any set $T$ with $S \subseteq T \subseteq \{1, \ldots, 2014\}$ is an element of $\mathcal{F}$. So $\mathcal{F} \supseteq \mathcal{F}_{S}$. Thus $\mathcal{F}=\mathcal{F}_{S}$.