The answer is $n=16$ and $n=17$. Call "initial set" the set of people with hats from the beginning. Label the people $0, 1, ..., n-1$. Strategy for $n=16$: .Let the person without a hat initially be number $x$. Then the assistant marks the person with number $15-x$ and let that person be $y$. Since $15$ is odd, $x\neq y$, so the assistant can do that. Then the magician comes and sees $y$ and the numbers $a$ and $b$ of the two people without hats. He checks which one of them is equal to $15-y$ and let that be $a$. So $b$ is the one who has taken off his hat and this way, the magician guesses it correctly. $\blacksquare$ Strategy for $n=17$: .Let the two people without a hat initially be $x$ and $y$. Then the assistant marks $z$ such that $2z\equiv x+y$ (mod $17$). Since $17$ is odd, there exists exactly one $z$ from $0$ to $16$ that satisfies the congruence. If $z$ is equal to one of $x$ and $y$, then this implies $x\equiv y$ (mod $17$), which is impossible. So the assistant is allowed to mark $z$. Now the magician comes and sees the marked number $z$ and the three people without hats, let them be $a$, $b$ and $c$. Let $z_a$, $z_b$ and $z_c$ be residues (mod $17$) such that $2z_a\equiv b+c$ (mod $17$), $2z_b\equiv a+c$ (mod $17$) and $2z_c\equiv a+b$ (mod $17$). Note that $z_a$, $z_b$ and $z_c$ are pairwise distinct because if two of them are equal, WLOG $z_a=z_b$, then $b+c\equiv2z_a\equiv2z_b\equiv a+c$ (mod $17$), which implies $a\equiv b$ (mod $17$), contradiction. The magician checks which one of $z_a$, $z_b$ and $z_c$ is equal to $z$, let it be $z_i$ where $i$ is $a$, $b$ or $c$. Then the person who took off his hat is $i$ and the magician guesses it correctly. $\blacksquare$ Why do all $n\geq 18$ not work: .Suppose there is a strategy for $n=18+k$ where $k$ is a nonnegative integer. For any set $S$ of $15$ people let $M(S)$ be the person the assistant has to mark according to the strategy. Claim: For any two sets of people $A$ and $B$ with $15$ elements such that $|A\cap B=14|$ we have $M(A)\neq M(B)$. Proof: Suppose the contrary. Let $A=\{x_1, x_2, ..., x_{13}, y, a\} $and $B=\{x_1, x_2, ..., x_{13}, y, b\}$, $M(A)=M(B)=y$. Let the remaining $k+2$ people be $l_1, l_2, ..., l_{k+2}$. Consider the scenario when the magician sees $a, b, l_1, l_2, ..., l_{k+2}$ with no hats and $y$ is marked. Then both $A$ and $B$ are possible initial sets and both $a$ and $b$ are possible people who had taken off their hats, so the magician fails. $\square$ That means that from any set $S$ of $16$ people the values of $M(T)$, where $T$ is a subset of $S$ with $15$ elements, are pairwise distinct. Let $P$ be the set of $15$-element subsets of $S$. Since there are $16$ people and $16$ different values, then the function $f: P \rightarrow S$ such that for all $T \in P$ $ f(T)=M(T)$ is a bijection. Fix a person $X$ and a set $Q$ of $18$ people including $X$ and consider the number $m$ of pairs of the form $(A, B)$ where $A$ is a subset of $Q$ of $16$ people including $X$ and $B$ is a subset of $A$ with $15$ people such that $M(B)=X$. On one hand, if we fix $A$, we have exactly $1$ possible option for $B$ because $f$ is bijective. So $m$ is the number of $16$-element subsets in $Q$ that include $X$, which is ${17\choose 15}=136$. On the other hand, if we fix $B$, we have $3$ options for $A$ - the union of $B$ and one of the remaining $3$ people in $Q$. That means $m$ is divisible by $3$, contradiction! Hence the magician fails and we are ready. $\blacksquare$