Solution:

The answer is yes. Take any arbitrary 15-element set (menu), and call a collection of subsets (meals) "valid" if the intersection of any two sets in the collection is again in the collection. Thus, the objective is to show that there exists a valid collection containing exactly 2001 sets. We will show, by downward induction, that there exists a valid collection with exactly $n$ sets for each $n$, $0 \leq n \leq 2^{15} = 32768$.

The base case $n = 2^{15}$ is clear: the collection of all subsets of the menu is certainly valid. For the induction step, suppose that there is a valid collection $C$ of $n$ subsets ($1 \leq n \leq 32768$); we will show that there is a valid collection of $n-1$ subsets. Choose a subset in $C$ of maximum possible size, and remove it; let $C'$ denote the remaining collection, so that it consists of $n-1$ subsets. We claim $C'$ is still valid. Indeed, if $S, T \in C'$, then the removed subset cannot be contained within either $S$ or $T$ (because of maximality), hence it certainly cannot equal their intersection. But $S \cap T$ was in $C$; hence, it is also in $C'$, as needed. Thus, the claim holds. Now, let $n = 2001$, and the problem is solved.