Let $S_n = a_1a_2\dots a_{15}\left (a_1^n+a_2^n+\dots+a_{15}^n\right )$. First, notice that $a^4 \equiv a^2 \equiv 1 \pmod 3$ if $(a,3)=1$, and that $\pmod 5$ if $(a, 5) = 1$ (by Fermat's little theorem). We show that $S_4$ is divisible by $15$ for all $a_i \in \mathbb N$. So, we should prove that $3| S_4$ and $5|S_4$. If there is some $i$ for which $a_i \equiv 0 \pmod 3$, then $3|S_4$. Otherwise we have \[S_4 \equiv \prod_{i=1}^{15} (a_i) \cdot \sum_{i=1}^{15} (a_i^4) \equiv \prod_{i=1}^{15} (a_i) \cdot 15 \equiv 0 \pmod 3.\] Exactly the same as above, one can show that $S_4 \equiv 0 \pmod 5$ and so $15 | S_4$. Now, we show that $n =4$ is the least positive integer for which $15 | S_n$. In other words, we show that $S_1$, $S_2$, and $S_3$ are not necessary divisible by $15$ for all $a_i$'s. And that's easy: Choose the counter example $(a_1, a_2, \ldots, a_{15}) = (\underbrace{1, 1, 1, \ldots, 1}_{\text{Fourteen 1's}}, 13)$.

Darn on phone but some motivation We are motivated by $\phi(15) = 8). Notice that if everything is not 3 or 5 mod 15 then when raised to the 8th power it is 1 mod 15. So we must consider if there exists numbers 3 or 5 mod 15 Clearly if both exist in the sequence then 15 divides a1. .. a15. If only a number 3 mod 15 but not 5 mod 15 exists in the sequence this means 3 divides a1 ... a15 and from phi(5) = 4 and 4 divides 8 we have that a1 ^8 + a2^8 +... a15^8 = 15 mod 5 = 0 mod 5 We use a similar argument for if there exists numbers 5 mod 15 but not 3 from phi(3) = 2. At this point we have not proven that 8 is the least possible n. Clearly it works as shown above, but what if something less works. We are motivated from phi(5) ,phi(3) dividing 4. Now notice 1^4, 2^4,4^4,7^4 are all 1 mod 15. Thus so are 8^4,11^4,13^4,14^4 This means we may apply the same argument for n=8 to n=4. It is easy to find counter examples for n=1. For n=2 just take 14 numbers in the sequence be 3 mod 15 and then 1 4 mod15 and for n=3 do the same Will latex when i get home