Suppose that $n \geq 2$. Consider the polynomial $P(t)=t(t-a_{1})(t-a_{2})...(t-a_{n})=t^{n+1}+c_{n}t^{n}+c_{n-1}t^{n-1}+...+c_{0}t$ Now, it's clear that $P(a_{i})=0$ for $i=1,2,...,n$, so: $0=P(a_{1})+P(a_{2})+...+P(a_{n}) = a_{1}^{n+1}+a_{2}^{n+1}+...+a_{n}^{n+1}+c_{n}(a_{1}^{n}+a_{2}^{n}+...+a_{n}^{n})+c_{n-1}(a_{1}^{n-1}+a_{2}^{n-1}+...+a_{n}^{n-1})+...+c_{0}(a_{1}+a_{2}+...+a_{n})$ From the given conditions it's clear that $c_{n}(a_{1}^{n}+a_{2}^{n}+...+a_{n}^{n})+c_{n-1}(a_{1}^{n-1}+a_{2}^{n-1}+...+a_{n}^{n-1})+...+c_{0}(a_{1}+a_{2}+...+a_{n})$ is divisible by $a_{1}+a_{2}+...+a_{n}$ so, we also have: $a_{1}+a_{2}+...+a_{n}|a_{1}^{n+1}+a_{2}^{n+1}+...+a_{n}^{n+1}$ Using this method it's very easy to prove inductively that: $a_{1}+a_{2}+...+a_{n}|a_{1}^{k}+a_{2}^{k}+...+a_{n}^{k}(*)$ for $k \in \mathbb{Z}_{+}$ (simply by considering the polynomial $P(t)=t^{k-n}(t-a_{1})(t-a_{2})...(t-a_{n})$ instead of $P(t)=t(t-a_{1})(t-a_{2})...(t-a_{n})$). Now, we will prove that $a_{1}+a_{2}+...+a_{n}|n(n-1)$. Let $a_{1}+a_{2}+...+a_{n}=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{m}^{\alpha_{m}}$ We shall show that $p_{i}^{\alpha_{i}}|n$ or $p_{i}^{\alpha_{i}}|n-1$ for $i=1,2,...,m$. Suppose that $a_{1}, a_{2}, ..., a_{n}$ are not divisible by $p_{i}$. Then $a_{j}^{\phi(p_{i}^{\alpha_{i}})}\equiv 1 \mod{p_{i}^{\alpha_{i}}}$, so plugging $k=\phi(p_{i}^{\alpha_{i}})$ in $(*)$ we get that: $0 \equiv a_{1}+...+a_{n}\equiv a_{1}^{\phi(p_{i}^{\alpha_{i}})}+a_{2}^{\phi(p_{i}^{\alpha_{i}})}+...+a_{n}^{\phi(p_{i}^{\alpha_{i}})}\equiv 1+1+...+1 \equiv n \mod{p_{i}^{\alpha_{i}}}$ Which means that $n$ is divisible by $p_{i}^{\alpha_{i}}$ Now, suppose that there is $j$ such that $1\leq j \leq n$ and $a_{j}$ is divisible by $p_{i}$. Then, $a_{j}^{\phi(p_{i}^{\alpha_{i}})}\equiv 0 \mod{p_{i}^{\alpha_{i}}}$ because it's easy to observe that $\phi(p^{t}) > t$. Also, there aren't any other numbers in the set $\{a_{1}, a_{2}, ..., a_{n}\}$ which are divisible by $p_{i}$ because they are pairwise coprime. Then, plugging again $k=\phi(p_{i}^{\alpha_{i}})$ in $(*)$ gives us: $0 \equiv a_{1}+a_{2}+...+a_{n}\equiv a_{1}^{\phi(p_{i}^{\alpha_{i}})}+a_{2}^{\phi(p_{i}^{\alpha_{i}})}+...+a_{n}^{\phi(p_{i}^{\alpha_{i}})}\equiv 1+1+...1+0+1...+1 \equiv n-1 \mod{p_{i}^{\alpha_{i}}}$ which means that $n-1$ is divisible by $p_{i}^{\alpha_{i}}$. So finally, we have that $a_{1}+a_{2}+...+a_{n}|n(n-1)$. Suppose that $a_{1}+a_{2}+...+a_{n}\not = n(n-1)$. Then it's obvious that $a_{1}+a_{2}+...+a_{n}\leq \frac{n(n-1)}{2}$ But $a_{1}, a_{2}, ..., a_{n}$ are pairwise distinct so: $a_{1}+a_{2}+...+a_{n}\geq 1+2+...+n= \frac{n(n+1)}{2}$ Contradiction. We are left with the case $a_{1}+a_{2}...+a_{n}=n(n-1)$ and we will prove that for $n \geq 2$ this is impossible. For $n \leq 4$ we can check it directly. For $n=5$ if one of $a_{1}, a_{2}, ... a_{5}$ would be divisible by $5$ then $a_{1}^{4}+...+a_{5}^{4}$ wouldn't be divisible by $5$. If not, then: $a_{1}+...+a_{5}\geq 1+2+3+7+11=24>20$. For $n=6$ and $n=7$ we get the same contradiction but with the divisibility by $3$ and $7$ respectively. And for $n \geq 8$: $a_{1}+a_{2}+...+a_{n}\geq 1+2+3+5+7+11+13+15+...+2n+1 = (1+3+5+7+9+...+2n+1)-7 = n^{2}-7$ It's clear that $n^{2}-7 > n^{2}-n$ for $n \geq 8$, so the proof is finished. Answer: there are no solutions if $n \geq 2$ and if $n=1$ $a_{1}$ is arbitrary.

TomciO wrote: Then $a_{j}^{\phi(p_{i}^{\alpha_{i}})}\equiv 1 \mod{p_{i}^{\alpha_{i}}}$, so plugging $k=\phi(p_{i}^{\alpha_{i}})$ in $(*)$ we get that: $0 \equiv a_{1}+...+a_{n}\equiv a_{1}^{\phi(p_{i}^{\alpha_{i}})}+a_{2}^{\phi(p_{i}^{\alpha_{i}})}+...+a_{n}^{\phi(p_{i}^{\alpha_{i}})}\equiv 1+1+...+1 \equiv n \mod{p_{i}^{\alpha_{i}}}$ Which means that $n$ is divisible by $p_{i}^{\alpha_{i}}$ Why $a_{1}+...+a_{n}\equiv a_{1}^{\phi(p_{i}^{\alpha_{i}})}+a_{2}^{\phi(p_{i}^{\alpha_{i}})}+...+a_{n}^{\phi(p_{i}^{\alpha_{i}})}$ ? Not $a_{1}\equiv a_{1}^{\phi(p_{i}^{\alpha_{i}})+1}$ ? thx..

Look at the relations: $p_{i}^{\alpha_{i}}|a_{1}+a_{2}+...+a_{n}$ and $a_{1}+a_{2}+...+a_{n}|a_{1}^{\phi(a_{1})}+a_{2}^{\phi(a_{2})}+...+a_{n}^{\phi(a_{n})}$.

It similar Hungari MO 1997 One idea

here is mine ... wlog $ a_1 < ... < a_n$ let $ s_i = \sum_{j = 1}^{n} a_j^i$ and $ P = \prod a_i$ and $ Q = \sum_{i = 1}^{n} \frac {P}{a_i}$ and it's obvious that $ \gcd(P,Q) = 1$ . we have $ a_1 + a_2 + ... + a_n \mid (a_1 + ... + a_n)(a_1^{n - 1} + ... + a_{n - 1})$ $ \implies a_1 + a_2 + ... + a_n \mid \sum_{i \neq j} a_ia_j(a_i^{n - 2} + a_j^{n - 2})$ $ \implies a_1 + a_2 + ... + a_n \mid \sum_{i \neq j} a_ia_j(s_{n - 2} - a_i^{n - 2} - a_j^{n - 2})$ $ \implies a_1 + a_2 + ... + a_n \mid \sum_{i,j,k \ \text{distinct}} a_ia_ja_k(a_i^{n - 3} + a_j^{n - 3} + a_k^{n - 3})$ $ \implies a_1 + a_2 + ... + a_n \mid \sum_{i,j,k \ \text{distinct}} a_ia_ja_k(s_{n - 3} - a_i^{n - 3} - a_j^{n - 3} - a_k^{n - 3})$ $ \implies a_1 + a_2 + ... + a_n \mid \sum_{i,j,k,t \ \text{distinct}} a_ia_ja_ka_t(a_i^{n - 4} + a_j^{n - 4} + a_k^{n - 4} + a_t^{n - 4})$ well... it's easy to understand that by doing the same thing again and again we'll finally reach this divisibility : $ a_1 + a_2 + ... + a_n \mid nP$ (1) now do exactly the same thing but this time by considering $ (a_1 + ... + a_n)(a_1^{n - 2} + ... + a_n^{n - 2})$ instead of $ (a_1 + ... + a_n)(a_1^{n - 1} + ... + a_{n - 1})$ at the beginning of the process ... this time we'll finally reach this divisibility : $ a_1 + a_2 + ... + a_n \mid (n - 1)Q$ (2) let me tell an example to clarify : if $ n = 4$ and we want to find the 4-th tuples (a,b,c,d) then ... $ a + b + c + d \mid (a + b + c + d)(a^3 + b^3 + c^3 + d^3)$ $ \implies a + b + c + d \mid ab(a^2 + b^2) + bc(b^2 + c^2) + cd(c^2 + d^2) + da(d^2 + a^2) + db(d^2 + b^2) + ac(a^2 + c^2)$ $ \implies a + b + c + d \mid ab(c^2 + d^2) + bc ( a^2 + d^2) + ... + ac ( b^2 + d^2)$ $ \implies a + b + c + d \mid abc(a + b + c) + bcd(b + c + d) + cda(c + d + a) + dab(d + a + b)$ $ \implies a + b + c + d \mid abc(d) + bcd(a) + cda(b) + dab(c) = 4abcd$ and also ... $ a + b + c + d \mid (a + b + c + d)(a^2 + b^2 + c^2 + d^2)$ $ \implies a + b + c + d \mid ab(a + b) + bc(b + c) + cd(c + d) + da(d + a) + db(d + b) + ac(a + c)$ $ \implies a + b + c + d \mid ab(c + d) + bc(a + d) + ... + ac(b + d) = 3 ( abc + bcd + cda + dab)$ back to the problem .. . from (1) and (2) we get ... $ a_1 + ... + a_n \mid \gcd(nP,(n - 1)Q)$ but we obviously have $ \gcd(nP,(n - 1)Q) \mid n(n - 1)$ so we must have $ a_1 + ... + a_n \mid n(n - 1)$ but since $ a_i$ are distinct we have $ a_1 + ... + a_n \ge 1 + 2 + ... + n = \frac {n(n + 1)}{2}$ and so if $ a_1 + ... + a_n < n(n - 1)$ then since $ a_1 + ... + a_n = \frac {n(n - 1)}{k}$ for some $ k$ so we get $ \frac {n(n + 1)}{2} \le a_1 + ... + a_n \le \frac {n(n - 1)}{2}$ which is a contradiction ... so we must have $ a_1 + ... + a_n = n(n - 1)$ we can easily prove that $ a_1 + ... + a_n \ge 1 + \sum_{i = 1}^{n - 1} p_{i}$ where $ p_i$ is the $ i - th$ prime and equality occurs iff $ a_1 = 1$ and $ a_i = p_{i - 1}$... now we can easily see that for $ n = 7$ we have $ 1 + \sum_{i = 1}^{6} p_{i} = 7 (7 - 1)$ but for $ n \ge 8$ we have $ 1 + \sum_{i = 1}^{n - 1} p_{i} > n(n - 1)$ meaning that we do not have any n tuple with the properties mentioned for $ n \ge 8$ ... it remains to check some trivial cases ...