Yes he can. Consider a 10-tuple ,call the changed pair (if it exists) bad and let $S$ be the set of the $100$ stones. Step_1 Let him consider $3$ disjoint $10$-tuples $A, B$ and $C$ and have him note down all the greater-smaller relations as they occur after a night has passed. Step_2 Consider all 10-tuples of this type: $2$ stones from each of $A$, $B$ and $C$ ($6$ in total) and a fixed $4$tuple from $S/ \{A,B,C\}$. Have him do this so that all of the $\binom{10}{2} \binom{10}{2} \binom{10}{2}$ $6$tuples occur in one such $10$-tuple. This means that one of these $10$-tuples must contain all $3$ bad pairs of $A$, $B$ and $C$, wlog $(a_1,a_2), (b_1,b_2), (c_1,c_2)$, respectively. with $a_1<a_2$ , $b_1<b_2$ ,$c_1<c_2$ in the actual weight catalog but with $a_1>a_2$ , $b_1>b_2$ , $c_1>c_2$ in the drums catalog and the one that the guy playing has noted down. Since , drum can only change $2$ of these $6$ stones one of these pairs remains for sure unchanged and hence $A$ has his notes contradicted. Which means there is a drum. If after the above procedure none of the notes he has written are contradicted, then the drum is not in his house.