Solution:

The answer is (D). I call $A$ the set of numbers that sum to $2024$ and whose product is given by the maximum possible value, that is $N$. I call "hole" a number $n$ that is not in $A$, but such that $A$ contains both a number greater than $n$ and a number smaller than $n$.

I prove that $A$ does not contain $1$: let $M$ be the maximum integer contained in $A$; if $A$ contained $1$, then the set $B$ obtained from $A$ by removing $1$ and adding $M+1$ would still respect all the conditions imposed by the problem, but the product of the numbers contained in it would be $N \cdot (M+1) > N$, contradicting the fact that $N$ was the maximum possible.

I prove by contradiction that $A$ has at most one hole: suppose that $a < b$, with $a, b \in A$ and such that $a+1$ and $b-1$ are two distinct holes; since $(a+1)(b-1) > a b$, the set $C$ obtained from $A$ by removing $a, b$ and adding $a+1$ and $b-1$ still respects all the hypotheses of the problem. Moreover, the product of the numbers contained in $C$ would be given by $N \frac{(a+1)(b-1)}{a b} > N$, contradicting the maximality of $N$.

Furthermore, $A$ cannot fail to contain $2$ while at the same time having a hole $n$ such that $n+2 \in A$. Indeed, if this were the case, the set $D$, obtained from $A$ by adding $2$ and $n$ and removing $n+2$, still respects the hypotheses of the problem. Once again, the product of the numbers contained in $D$ is given by $N \frac{2 n}{n+2} > N$, which implies that $N$ was not the maximum possible.

The only way to write $2024$ as a sum of distinct numbers respecting the previous conditions is by using the numbers from $2$ to $64$, skipping $55$. It follows that $N = \frac{64!}{55}$, which ends with $13$ digits '0'. To count the number of trailing zero digits it is enough to count the number of factors $2$ and the number of factors $5$ that appear in the factorization of $N$, and then take the minimum of these two quantities. It is easy to verify that there are fewer factors $5$ than factors $2$; finally, to compute the number of factors $5$, one can observe that these are the same as those of the product $5 \cdot 10 \cdot 15 \cdots 50 \cdot 60$: this product has $1+1+1+1+2+1+1+1+1+2+1=13$ factors $5$ (indeed every multiple of $5$ between $5$ and $60$ contains only one factor $5$, except for $25$ and $50$ which contain $2$ each).