Easy for N6 in my opinion. Proving $n=1, 2$ fails is not too difficult, and proving $n=3$ can be shown by simple vieta jumping.

Really nice problem.

The smallest number is $n=3$. (Showing $n\geq 3$ isn't very hard so I'll omit soz) To show $n=3$ works, let's consider the construction \[ab+\frac{a}{a+b}+\frac{b}{a+b}\in \mathbb{Z}\]\[\frac{1}{ab}+\frac{a+b}{a}+\frac{a+b}{b}\in \mathbb{Z}.\]The first one is trivially true; the second one would be true if $ab\mid (a+b)^2+1$. Assume $a,b$ coprime. Then you would need $a\mid b^2+1$ and $b\mid a^2+1$. The pair $(5,13)$ is a solution to this. Now for a certain $(a,b)$, it is not very hard to verify that $(a,\frac{a^2+1}{b})$ is also a solution, and so \[\frac{(\frac{a^2+1}{b})^2+1}{a},\frac{a^2+1}{b}\]is also a solution. All that remains is to show that if $a>b$, then this produces arbitrarily large solutions, which are then of course distinct.

The answer is $n=3$. First, $n=1$ clearly fails. We show $n=2$ fails: if $a+b = p$ and $\frac{1}{a}+\frac{1}{b} = q$ for integers $p$ and $q$. Let $a = x/y$ with $\gcd(x,y)=1$; then \[ q = \frac{1}{a} + \frac{1}{p-a} = \frac{p}{a(p-a)} = \frac{py^2}{x(py-x)}. \]Then $x \mid p$, so we may write $p=xk$ and obtain \[ q = \frac{xk \cdot y^2}{x^2(ky-1)} = \frac{ky^2}{x(ky-1)}. \]As $\gcd(ky-1, ky^2) = 1$, this forces $ky-1=1$, or $ky=2$. If $y=1$ then $a \in {\mathbb Z}$, so $b \in {\mathbb Z}$, and so either $a=b=1$ or $a=b=2$. If $y=2$, then $a$ and $b$ are both half-integers, and so we conclude $a=b=\frac{1}{2}$. Now to show $n=3$ works, we take a triple of the form \[ \left( \frac{1}{1+x+y}, \frac{x}{1+x+y}, \frac{y}{1+x+y} \right) \]where $x$, $y$, $z$ are positive integers (in fact if we pick $x,y,z \in {\mathbb Q}$ this is WLOG). Then it suffices that \[ \frac{1+y}{x} + \frac{1+x}{y} \in {\mathbb Z} \]which is a famous MOP 2007 problem solved by Vieta jumping (there are in fact infinitely many with $\frac{1+y}{x} + \frac{1+x}{y} = 3$).

The answer is $n = 3$. First we discard the case $n \leq 2$. For $n = 1$ clearly only $a_1 = 1$ works. For $n = 2$ let $a_1 = \frac{a}{b}$ and $a_2 = \frac{c}{d}$ with $(a, b) = (c, d) = 1$. The conditions then imply that $bd$ divides $ad + bc$ and $ac$ divides $bc + ad$. Then $d$ divides $bc$ and because $(c, d) = 1$ it follows that $d \mid b$. Analogously $b \mid d$, $a \mid c$ and $c \mid a$, so in fact $a_1 = a_2$, so $2a_1$ and $\frac{2}{a_1}$ are both integers giving the pairs $(1/2, 1/2), (1, 1)$ and $(2, 2)$. Now we prove that $n = 3$ works. We will in fact prove that there exist infinitely many triples of positive rationals $(a, b, c)$ with $a + b + c = 1$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ an integer (this actually follows if the original problem is true, so it can't make the problem false and it adds a nice bit of structure). Let $a = \frac{x}{x + y + z}$ and the analogous where $x, y, z$ are positive integers. Then $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{x + y + z}{x} + \frac{x + y + z}{y} + \frac{x + y + z}{z}$$ So we want $\frac{y + z}{x} + \frac{z + x}{y} + \frac{x + y}{z}$ to be an integer. We now impose $x = 1$, so we want $\frac{z + 1}{y} + \frac{y + 1}{z} = \frac{y^2 + z^2 + y + z}{yz}$ to be an integer. Assume that for some positive integers $y, z, k$ this ratio is equal to $k$, with WLOG $z \geq y$. Then $$y^2 + z^2 + y + z = kyz \implies y^2 - (kz - 1)y + z^2 + z = 0$$ Seen as a quadratic equation with variable $y$, its other root is $kz - y - 1 \geq (k - 1)z - 1$, so if $k \geq 4$ then this other root is an integer greater than $z$, so this process gives us a new pair $(z, kz - y - 1)$ that gives the ratio $k$ from the pair $(y, z)$. Because the pair $(1, 1)$ satisfies the desired equality with $k = 4$, we may construct infinitely many triplets from this, as desired.

Nevermind.

fattypiggy123 wrote: Some background on this problem: some years ago a student (known for fake-solving) came to us with this problem with the claim that there were finitely many solutions for all $n$. Of course his solution was wrong but the problem was interesting and we solved it ourselves and proposed it later on. Oneplusone wanted the problem to have a clear goal (either show $n = 3$ had infinitely many solutions for show there exists such a $n$) but I chose this wording since it was the situation we were in when we first attempted it and it wasn't exactly clear what the answer would be. So, you gave that student no credit? Why not say he was a co proposer?

SHARKYKESA wrote: Find the smallest positive integer $n$ or show no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $(a_1, a_2, \ldots, a_n)$ such that both $$a_1+a_2+\dots +a_n \quad \text{and} \quad \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}$$are integers. Solution. The answer is $n=3.$ It is easy to see that $n=1$ and $n=2$ doesn't work. We will show that $n=3$ works. Take \begin{align*} a_1= \frac{xy+y^2}{xy-1},\qquad a_2 =\frac{xy+x^2}{xy-1},\qquad a_3 = xy \end{align*}where $x$ and $y$ are positive integers. It is easy to verify that $\tfrac{1}{a_1} + \tfrac{1}{a_2} + \tfrac{1}{a_3}=1$. Note that \[a_1+a_2+a_3=xy+\frac{2xy+x^2+y^2}{xy-1}=xy+2+\frac{x^2+y^2+2}{xy-1}.\]Hence we want infinitely many $(x,y)\in\mathbb N^2$ such that $xy-1\mid x^2+y^2+2.$ Define the sequences $\langle x_n\rangle$ and $\langle y_n\rangle$ such that \begin{align*} (x_1,y_1)&=(2,1)\\ (x_{n+1},y_{n+1})&=\begin{cases}(7y_n-x_n,y_n), \text{ if }7y_n-x_n>x_n \\ (x_n,7x_n-y_n),\text{ otherwise }\end{cases} \end{align*}It is easy to check by induction that all the pairs $(x_n,y_n)$ work. Also, the value $x_n+y_n$ is strictly increasing. Thus, we get infinitely many pairs $(x,y).$ And we are done. $\blacksquare$

simple sol: $abc|ab+bc+ca \implies abc|c^2(a+b) \implies ab|c(a+b)$ which easily we can choose $c$ which such in this.

arzhang2001 wrote: simple sol: $abc|ab+bc+ca \implies abc|c^2(a+b) \implies ab|c(a+b)$ which easily we can choose $c$ which such in this. You take $a,b,c$ integer? If not, you must prove $a + b + c$ integer. Nevertheless, choosing $ab | c(a+b)$ doesn't ensure $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer. Take an example (1,1,2020)

no my friend. we choose$a,b$ such $a+b$ be integer and $c$ is a integer. good luck

arzhang2001 wrote: no my freind. we choose$a,b$ such $a+b$ be integer and $c$ is a integer. good luck So basically you choose any $a,b$ such that $a+b$ is an integer and $c$ is an integer such that $ab |c(a+b)$. I don't really see why this imply $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= \frac{ac + bc + ab}{abc} = \frac{1}{c} \left( \frac{ac + bc}{ab} + 1 \right) $ must be an integer. You only take such that $\frac{ac + bc}{ab}$ is an integer

if $\frac{ab+ba+ca}{abc}$ be integer then should $\frac{(ab+bc+ca)c-abc}{abc}$ also be integer which means:$abc|c^2(a+b) \implies ab|c(a+b)$

arzhang2001 wrote: if $\frac{ab+ba+ca}{abc}$ be integer then should $\frac{(ab+bc+ca)c-abc}{abc}$ also be integer which means:$abc|c^2(a+b) \implies ab|c(a+b)$ Reverse direction doesn't hold lol, we want $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ to be an integer. You said that any $a,b,c$ such that $ab | c(a+b)$ should give us $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer. This isn't the case $(1,1,2020)$.

because from this ($\frac{(ab+bc+ca)c-abc}{abc}$ be integer) we cant get( $\frac{ab+ba+ca}{abc}$is integer) assumption which i said before is necessary but not enough you can choose $a,b$ in such condition: $\frac{a+b}{ab}=\frac{ck-1}{c} $(k is integer). which you can reach to this because : if you fix $a+b=t$ then you can show that every rational numbers which bigger than $\frac{4}{t}$ with this form$a+b/ab$ according your last post which yo said if you put a,b,c on this you get : $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= \frac{ac + bc + ab}{abc} = \frac{1}{c} \left( \frac{ac + bc}{ab} + 1 \right)= $integer. note: forget $r$

arzhang2001 wrote: because from this ($\frac{(ab+bc+ca)c-abc}{abc}$ be integer) we cant get( $\frac{ab+ba+ca}{abc}$is integer) assumption which i said before is necessary but not enough you can choose $a,b$ in such condition: $\frac{a+b}{ab}=\frac{ck-1}{c} =\frac{r-1}{r}$(k is integer). which you can reach to this because : if you fix $a+b=t$ then you can show that every rational numbers which bigger than $\frac{4}{t}$ with this form$a+b/ab$ according your last post which yo said if you put a,b,c on this you get : $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= \frac{ac + bc + ab}{abc} = \frac{1}{c} \left( \frac{ac + bc}{ab} + 1 \right)= $integer. note: thanks you for help me complete my proof Looks like a fakesolve to me. 1. What's the use of $r$. 2. You claim that every rational number greater than $4/t$ can be written as $\frac{a + b}{ab}$. This is obviously false. 3. All of your claims seems too flawty . You are asked to show the existence of infinitely many triples yet you give so much holes in your argument. This would be 0 in an essay i would say.

arzhang2001 wrote: because from this ($\frac{(ab+bc+ca)c-abc}{abc}$ be integer) we cant get( $\frac{ab+ba+ca}{abc}$is integer) assumption which i said before is necessary but not enough you can choose $a,b$ in such condition: $\frac{a+b}{ab}=\frac{ck-1}{c} =\frac{r-1}{r}$(k is integer). which you can reach to this because : if you fix $a+b=t$ then you can show that every rational numbers which bigger than $\frac{4}{t}$ with this form$a+b/ab$ according your last post which yo said if you put a,b,c on this you get : $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= \frac{ac + bc + ab}{abc} = \frac{1}{c} \left( \frac{ac + bc}{ab} + 1 \right)= $integer. note: thanks you for help me complete my proof Sorry, but I don't get your wording/point. Can you please reexplain it clearly? (What is $r$?, how can you ensure that there are infinite number of triplets $(a,b,c)$ which fits the bill? Just explain it from the beginning.)

Mathological03 wrote: arzhang2001 wrote: because from this ($\frac{(ab+bc+ca)c-abc}{abc}$ be integer) we cant get( $\frac{ab+ba+ca}{abc}$is integer) assumption which i said before is necessary but not enough you can choose $a,b$ in such condition: $\frac{a+b}{ab}=\frac{ck-1}{c} =\frac{r-1}{r}$(k is integer). which you can reach to this because : if you fix $a+b=t$ then you can show that every rational numbers which bigger than $\frac{4}{t}$ with this form$a+b/ab$ according your last post which yo said if you put a,b,c on this you get : $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= \frac{ac + bc + ab}{abc} = \frac{1}{c} \left( \frac{ac + bc}{ab} + 1 \right)= $integer. note: thanks you for help me complete my proof Looks like a fakesolve to me. 1. What's the use of $r$. 2. You claim that every rational number greater than $4/t$ can be written as $\frac{a + b}{ab}$. This is obviously false. 3. All of your claims seems too flawty . You are asked to show the existence of infinitely many triples yet you give so much holes in your argument. This would be 0 in an essay i would say. 1. What's the use of $r$? arbitrary number 2.You claim that every rational number greater than $4/t$ can be written as $\frac{a + b}{ab}$. This is obviously false. i going to prove it true.(i wrongly say fix $a+b$) $\frac{a+b}{ab}=\frac{t}{ab}=\frac{p}{q} \implies $$ ab=a(t-a)=\frac{p}{qt}$ which $p ,q$ arbitrary . you can choose $t$ which this quadratic equation have root. 3. never mind. i tired to enplane same things to anther person. note: i know maybe my proof have gap . honestly i have ADHD or Attention Deficit Hyperactivity Disorder which when i forget to eat my drugs i have many mistake like computing or didn't see part of problem or... but i sure about my idea.my proof gaps not hurt my idea. @below: hi . 1.actually $r$ not important for solution .you suppose $c$ fixed number which in this case you trust but $c$ not fixed. if you see my post i wrote: forget $r$. 2.you trust but as i said before its not hurt proof. you can choose $\frac{ck-1}{c}$ such in $ ab=a(t-a)=\frac{p}{qt}$. 3.i sorry to wrote simple proof .but as you can see in IMO problems most of them have simple & short proof (may p6 or p3) but this not necessarily mean they easy. have nice day guys.

Okay I might be too harsh. I just don't really like untidy ideas or something (ocdc) arzhang2001 wrote: 2.You claim that every rational number greater than $4/t$ can be written as $\frac{a + b}{ab}$. This is obviously false. i going to prove it true.(i wrongly say fix $a+b$) $\frac{a+b}{ab}=\frac{t}{ab}=\frac{p}{q} \implies $$ ab=a(t-a)=\frac{p}{qt}$ which $p ,q$ arbitrary . you can choose $t$ which this quadratic equation have root. 1. arbitrary number? Must there exists one? I dont see any use or $r$ in the following arguments. 2. $ab = a(t - a) = \frac{p}{qt}$ does has a root, but not necessarily "rational" root. 3. I'm sorry for your ADHD. But instead of saying "easy solution" for an N6 problem, please just state I have an idea or something, if you thinj that would work but not sure about the gap things. It seems annoying for others.

Inconsistent wrote: It's trivial by bounding that $n = 1, 2$ fail. For $n = 3$, simply take for nonnegative integers $k$, $$(a_1, a_2, a_3) = \left(\frac{2}{((2+\sqrt{3})^k+(2-\sqrt{3})^k)^2+2}, \frac{((2+\sqrt{3})^k+(2-\sqrt{3})^k+\frac{(2+\sqrt{3})^k-(2-\sqrt{3})^k}{\sqrt{3}}) \cdot ((2+\sqrt{3})^k+(2-\sqrt{3})^k)}{2((2+\sqrt{3})^k+(2-\sqrt{3})^k)^2+4}, \frac{((2+\sqrt{3})^k+(2-\sqrt{3})^k-\frac{(2+\sqrt{3})^k-(2-\sqrt{3})^k}{\sqrt{3}})\cdot ((2+\sqrt{3})^k+(2-\sqrt{3})^k)}{2((2+\sqrt{3})^k+(2-\sqrt{3})^k)^2+4}\right)$$ The verification is left as an exercise for the reader. What is the exact motivation behind this? Just wanted to know out of interest.

Sross314 wrote:

What is the exact motivation behind this? Just wanted to know out of interest. Same as the other solutions, more or less. The claim is that we can find $\frac{1}{1+x+y}, \frac{x}{1+x+y}, \frac{y}{1+x+y}$ to be a solution, and we achieve this by reducing special cases (for example, I choose $x+y = 2\gcd(x, y)^2$ as my special case) into a generalized pell equation on which I impose a few parity restrictions to get a standard pell equation so that I can write solutions in terms of the fundamental solution (specifically for this one it was $x^2-3y^2=1$). This is essentially equivalent to the Vieta jumping solutions, see 2019 ISL N8 for details on how to do this interchanging.

For $n=1$, $a_1$ and $\frac1{a_1}$ must be integers, so $a_1=1$ is the only solution. If $n=2$, let $a_1=\frac ab$ and $a_2=\frac cd$ with $\gcd(a,b)=1$ and $\gcd(c,d)=1$ ($a,b,c,d$ are integers). We require: $$\frac ab+\frac cd=\frac{ad+bc}{bd}$$to be an integer. From this we obtain $b\mid ad$, so $b\mid d$ and similarly $d\mid b$. This means that $b=d$. From $\frac ba+\frac dc=\frac{ad+bc}{ac}$ being an integer, we also need $a=c$, so $a_1=a_2$. But then $2a_1$ and $\frac1{2a_1}$ are integers, which only has finitely many solutions. We claim that there are infinitely many $(a,b)\in\mathbb Z^2$ such that: $$\frac a{a+b+1}+\frac b{a+b+1}+\frac1{a+b+1}$$and $$\frac{a+b+1}a+\frac{a+b+1}b+a+b+1$$are integers. The first one is obviously true; it's equal to $1$. For the second we require $ab\mid a^2+b^2+a+b$. Call such a pair of positive integers $(a,b)$ good. If $(a,b)$ is good, then so are $(b,a)$ and $\left(\frac{b^2+b}a,b\right)$ (note that $a\mid b^2+b$ if $(a,b)$ is good). Assume FTSOC that there is a pair $(a,b)$ with $a+b$ maximal such that $(a,b)$ is still good. If $a\le b$ then consider the good pair $\left(\frac{b^2+b}a,b\right)$. We have: $$\frac{b^2+b}a+b>\frac{b^2}a+b\ge\frac{a^2}a+b=a+b$$so $a+b$ is not maximal. If $a\ge b$ then $(b,a)$ is good, so it goes back to the previous paragraph. Either way, we must have infinitely many good pairs, so $n=\boxed3$ works.

idk if works Note that the only solution for $n=1$ is $a_1=1$ so there are not infinitely many. Now let $n=2.$ We have $a_1+a_2=c$ and $\frac{1}{a_1}+\frac{1}{a_2}=d$ so we can divide both $a_1$ and $a_2$ by $c$ to get a new set of solutions $a_1+a_2=1$ and $\frac{1}{a_1}+\frac{1}{a_2}=cd$ where $cd$ is an integer. We have $\frac{1}{a_1a_2}$ must be an integer, so $a_1=a_2=\frac{1}{2}$. Thus, $a_1=a_2$ in every solution. $2a_1$ and $\frac{2}{a_1}$ are integers so there are finitely many solutions. Now let $n=3.$ We may substitute: $a_1=\frac{b_1}{b_1+b_2+b_3},a_2=\frac{b_2}{b_1+b_2+b_3},a_3=\frac{b_3}{b_1+b_2+b_3}.$ We know that $a_1+a_2+a_3$ becomes $1$ and $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}$ becomes $\left(b_1+b_2+b_3\right)\left(\frac{1}{b_1}+\frac{1}{b_2}+\frac{1}{b_3}\right).$ Now, let $b_1=1$ so $(1+b_2+b_3)\left(1+\frac{1}{b_2}+\frac{1}{b_3}\right)$ is an integer. Thus, we want $\frac{1+b_3}{b_2}+\frac{1+b_2}{b_3}$ integer. Let $(b_2,b_3)$ be a valid pair then $b_3^2+b_2^2+b_3+b_2=kb_2b_3$ so $b_3^2+(1-kb_2)b_3+b_2^2-b_2$ has integer solutions. We have $(b_2,kb_2-1-b_3)$ is a valid solution, and bounding confirms: if $b_2>b_3$ then $kb_2-1-b_3>b_2$ as long as $k>3.$ This, it comes down to finding a valid $k.$ We have $\frac{1+3}{2}+\frac{1+2}{3}=3$ so we're done.

we claim the answer is $n=3$ $n=1$ fails for very obvious reasons so let $n=2$. Write $a_1=\frac{p}{q}$ and $a_2=\frac{r}{s}$ where $\gcd(p,q)=\gcd(r,s)=1$. We have \[ \frac{p}{q}+\frac{r}{s} = \frac{ps+rq}{pq} \in \mathbb{N} \text{ and } \frac{q}{p}+\frac{s}{r} = \frac{ps+rq}{rs} \in \mathbb{N}\]So, $p$, $q$, $r$, and $s$ all divide $ps+rq$. \[ p \vert ps+rq \implies p \vert rq \implies p \vert r \text{ but } r \vert ps+rq \implies p \vert ps \implies r \vert p\]So, $p=r$. By the same logic, $q=s$. So, $a_1=a_2$. So, \[ \frac{1}{a_1}+\frac{1}{a_2} = \frac{2}{a_1} \in \mathbb{N} \implies a_1=\frac{2}{k} \]for some $k \in \mathbb{N}$. But, \[ a_1+a_2 = 2a_1 = \frac{4}{k} \in \mathbb{N}\]So, there is only finitely many choices of $k$. Now, we will show $n=3$ works. Let \[ (a_1,a_2,a_3) = \left (\frac{1}{m+n+1}, \frac{m}{m+n+1}, \frac{n}{m+n+1} \right ) \]This boils down to proving \[ \frac{m+1}{n} + \frac{n+1}{m} \in \mathbb{N} \]for infinitely many pairs $(m,n)$. We will actually show that \[ \frac{m+1}{n}+\frac{n+1}{m} =4 \]infinitely often. Indeed, notice that $(m,n)=(1,1)$ works. Now assume that there exists some pair $(m,n)$ which satisfies the equality and $m+n$ is maximized. Assume wlog $m<n$ \[ \frac{m+1}{n}+\frac{n+1}{m} =4 \implies m^2-(4n-1)m+n^2+n=0 \]which is a quadratic in terms of $m$. So, let $m_1$ be the other root. First, notice that $m+m_1 = 4n-1 \implies m_1 \in \mathbb{Z}$. By maximality, we know that $m_1 \leq m$. So, \[ n^2+n = mm_1 \leq m^2 \implies n<m \]Contradiction.

The answer is $\boxed{n=3}$. To prove this works, we first show that $n=2$ doesn't work and then present and prove a construction for the $n=3$ case. $n=2$: Suppose the rational numbers are $\frac{a}{b}$ and $\frac{c}{d}$ in reduced form. Then $a_1+a_2=\frac{ad+bc}{bd}$ and $\frac{1}{a_1}+\frac{1}{a_2}=\frac{ad+bc}{ac}$. However, since $ac$ and $bd$ are relatively prime and not $1$, the two quantities cannot be equal. Thus, the $n=2$ case yields no solutions. $n=3$: We present the construction $\bigg(\frac{1}{m+n+1}, \frac{m}{m+n+1}, \frac{n}{m+n+1}\bigg)$, for all positive integers $m$, $n$ with $\frac{m+1}{n}+\frac{n+1}{m}=4$. To show that this construction works, we prove the following claim. Claim: The are infinitely many pairs of positive integers $(m, n)$ with $\frac{m+1}{n}+\frac{n+1}{m}=4$. Proof: Rewriting, this is $m^2-(4n-1)m+(n^2+n)=0$. Now, for each solution $(a, b)$, define a move by a transformation taking $(a, b)$ to $(\max(a, b), 4\max(a, b)-\min(a, b)-1)$. Notice that the maximum of the numbers in the pair increases with each move if we start with the solution $(1, 1)$, and that each pair produced is also a solution to the original Diophantine equation. Thus, iteratively applying moves to $(1, 1)$ generates an infinitude of such pairs, as claimed. $\square$ Therefore, there are infinitely many solutions in this construction, so $n=3$ works, and we are done.

The answer is $n=3$. Claim. There are infinitely many pairs $(m, n)$ with $$\frac{m+1}n+\frac{n+1}m$$a positive integer. Proof. This is well-known. See HMMT 2017 for instance. First, I show the $n=2$ case does not work. WLOG we may set $a_1+a_2 = 1$, so set $a_1 = \frac ab$. The condition implies $\frac{b^2}{a(b-a)}$ is an integer, but as $(a, b) = 1$, this obviously cannot yield infinitely many solutions. For the $n=3$ case, use the construction $$\left(\frac 1{m+n+1}, \frac m{m+n+1}, \frac n{m+n+1}\right).$$ Remark. [On Construction] The construction is actually quite natural, coming from fixing $a_1+a_2+a_3 = 1$ (as in the $n=2$ proof), parametrizing accordingly and forcing one variable to equal $1$ to make our life easier.

n=1 doesn't work, and n=2 with the two terms as a/b,c/d and gcd(a,b)=gcd(c,d)=1 gives $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\in\mathbb{Z}\implies b\mid d,d\mid b\implies b=d;\quad\frac ba+\frac dc\in\mathbb{Z}\implies a=c\implies 2a_1,\frac2{a_1}\in\mathbb{Z},$$which obviously has finite solutions. We claim the answer is n=3, constructed by $(a_1,a_2,a_3) = \left (\frac{1}{m+n+1}, \frac{m}{m+n+1}, \frac{n}{m+n+1} \right)$, since there are infinite solutions to $\frac{m+1}{n} + \frac{n+1}{m} \in \mathbb{Z}$ by Vieta jumping $(a,b)\leftrightarrow(\frac{b^2+b}a,b)$.

n6?? First, we prove $n=2$ doesn't work. Set $a+b=p, ab = \frac{p}{q}$ for positive integers $p,q$. Then $p^2 - \frac{4p}{q} = k^2$ for $k\in \mathbb Q$. We can rewrite this as $p^2q^2 - 4pq = k'^2$ for $k\in \mathbb N$. Treating this as a quadratic in $pq$, the discriminant must be a perfect square, or $4(k'^2 + 4)$, which obviously is only satisfied if $k' = 0$. For $n=3$, consider $\left(\frac{1}{1+p+q},\frac{p}{1+p+q}, \frac{q}{1+p+q}\right)$. It is equivalent to prove there are infinitely many $p,q$ such that $\frac{p+1}{q} + \frac{q+1}{p}$ is an integer, but this is trivial by Vieta Jumping.

Answer: $n=3$. Obviously $n \neq 1$. Assume $n=2$. Then there are infinitely many pairs of positive rationals $(x, y)$ such that $x+y$ and $\frac{1}{x} + \frac{1}{y}$ are both integers. Let $(x, y)$ be a pair of positive rationals such that $x + y$ and $\frac{1}{x} + \frac{1}{y}$ are both integers. Let $x = \frac{a}{b}, y = \frac{c}{d}$. Then it's not hard to see that $a=c$ and $b=d$, so $\frac{2a}{b}$ and $\frac{2b}{a}$ are integers. Thus there are finitely many pairs of positive rationals $(x, y)$ such that $x + y$ and $\frac{1}{x} + \frac{1}{y}$ are both integers. Now assume $n=3$. We'll prove that there are infinitely many positive rationals $x, y, z$ such that $x + y + z$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ are both integers. Take $(x, y, z) = (\frac{1}{1+a+b}, \frac{a}{1+a+b}, \frac{b}{1+a+b})$. Then obviously $x+y+z = 1$ and we only have to prove that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is an integer, which is equivalent to $\frac{a+1}{b} + \frac{b+1}{a}$ is an integer. Thus we only need to consider the following claim: Claim: There are infinitely many pairs of positive integers $(m, n)$ such that $\frac{m+1}{n} + \frac{n+1}{m}$ is an integer. Proof. Let $(m, n)$ be a pair of positive integers such that $\frac{m+1}{n} + \frac{n+1}{m}$ is an integer. Let $k = \frac{m+1}{n} + \frac{n+1}{m}$. Then we have $kmn = m^2 + n^2 + m + n$ and assume $n \ge m$. Then we have $m^2 + m(1 - kn) + (n^2 + n) = 0$ and let $m_1$ be a root of $x^2 + x(1 - kn) + (n^2 + n)$ other than $m$. Then we have $m + m_1 = (kn - 1)$ and $mm_1 = n^2 + n$, so $m_1$ is a positive integer. And note that $m_1 = \frac{n^2 + n}{m} > \frac{n^2}{m} \ge \frac{m^2}{m} = m$ , so if $(n, m)$ is a solution, then $(\frac{n^2 + n}{m}, n)$ is also solution and $\frac{n^2 + n}{m} > m$ implies we can generate infinitely many solutions from $(n, m)$. Take $(n, m) = (2, 1)$ and we can generate infinitely many solutions from $(2, 1)$. $\square$ Thus by claim, we're done. $\blacksquare$

I claim that the smallest such $n$ is $\boxed{n = 3}$. We first show that $n = 1, 2$ do not work for an infinite number of $n$-tuples: Case 1: ($n = 1$). Then $1/x \in \mathbb{Z} \implies x \vert 1 \implies x =1$, only. $\square$ Case 2: ($n = 2$). Then let $x = a/b, y = c/d$. We then have that as \begin{align*} d \vert ad + bc &\implies d \vert bc \implies d \vert b \\ b \vert ad + bc &\implies b \vert ad \implies b \vert d \\ c \vert ad + bc &\implies c \vert ad \implies c \vert a \\ a \vert ad + bc &\implies a \vert bc \implies a \vert c, \end{align*}that $b = d, a = c$. Hence $x + y = 2x$ and $1/x + 1/y = 2/x$. Then $2/x = 4/(2x) \implies x \in \{1/2, 1, 2\}$, which yields a finite number of solutions. $\square$ Claim: $n = 3$ is the smallest such value of $n$ providing an infinite number of $n$-tuples. Proof: Let $x, y, z \in \mathbb{Q}_{>0}$. Then we will construct valid solutions for when $(x, y, z) = \left(\frac{a}{a + b + c}, \frac{b}{a + b + c}, \frac{c}{a + b + c} \right)$, (i.e., when $x + y + z = 1$). It suffices to show there are an infinite number of solutions for $\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c} \in \mathbb{Z}$. Without loss of generality, assume $a < b < c$. Fix $a = 1$: Then we claim there are an infinite number of solutions to the equation $\frac{b^2 + c^2 + b + c}{bc} = 3$. First observe the smallest nontrivial solution $(b, c) = (2, 3)$. Let $f(t) = t^2 - t(3c - 1) + (c^2 + c)$. Then $b, b_0$ are roots, and $b + b_0 = c^2 + c \implies b_0 \in \mathbb{Z}$, and $bb_0 = c^2 + c$. Now observe \[b_0 = \frac{c^2 + c}{b} > \frac{c^2 + c}{c} > c.\]Then $b_0 + c > b + c$, so that there exists an infinite number of larger solutions upon reiterating this algorithm, as was needed to show. $\blacksquare$

Well this is weird showing that $n=1,2$ is not possible is easy for $n=3$ Choose $a_1=\frac{x}{x+y+z},a_2=\frac{y}{x+y+z},a_3=\frac{z}{x+y+z}$ Basically we want $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ integer we just let $x,y,z \in \mathbb{Z^+}$ . Now to show there are infinitely many $(x,y,z)$ to make that integer [also making $a_1,a_2,a_3$ distinct] Now set $a_3=a_1a_2$ , NOW this is an old HMMT problem which was solved by infinite descent.

$n=1$ is immediate. For $n=2$ note that if $a+b=p$ and $\frac 1a+\frac 1b=q$ then $q=\frac 1a+\frac 1{p-a}=\frac p{a(p-a)}$. Let $a=\frac mn$, then $q=\frac {pn^2}{m(np-m)}$ so we want $m\mid pn^2$. Since $m,n$ are coprime we want $m\mid p$ and similarly, we want $(np-m)\mid pn^2$. If there is a prime $r$ such that $r\mid np-m$ and $r\mid n$ then $r\mid m$, which is impossible as $m,n$ are coprime. Hence, we simply have $np-m\mid p$ which means $np-m\leq p$ and hence $m\geq (n-1)p$. But we just showed that $m\mid p$, so $n=2$ is forced as $n=1$ is impossible or else $a,b$ are integers. For $n=2$, we have $m=p$ forced and hence $pq=4$, and this leaves us with only finite solutions. For $n=3$, consider $a_1=ab$, $a_2=\frac a{a+b}$ and $a_3=\frac b{a+b}$ where $a,b$ are coprime integers. $a_1+a_2+a_3=ab+1$ is integer, and we show that $\frac 1{a_1}+\frac 1{a_2}+\frac1{a_3}$ is integer. We show that there are infinitely many pairs $(a,b)$ such that $ab\mid (a+b)^2+1$ or $ab\mid a^2+b^2+1$. This is a famous problem on Vieta Jumping that there are infinitely many pairs $(a,b)$ such that $\frac{a^2+b^2+1}{ab}$ is integer, and that integer has to be 3. Let $a^2+b^2+1=3ab$, or basically $a^2-3ab+b^2+1=0$. So, if $(a,b)$ is a solution then $(a,3b-a)$ is a solution and so is $(3a-b,b)$. We get many solutions like this: \[(1,1)\longrightarrow(1,2)\longrightarrow(5,2)\longrightarrow(5,13)\longrightarrow(34,13)\longrightarrow\cdots\]So, $n=3$ is the smallest such positive integer. $\blacksquare$

Answer $n=3$. $n=1$ clearly fails and for $n=2$ note that $(\tfrac1a+\tfrac1b)(a+b)=2+\tfrac ab+\tfrac ba$ is an integer, but taking $\nu_p$ with a prime dividing $\tfrac ab$ or $\tfrac ba$ implies noninteger, so $a=b$ and we can see $a=b=\tfrac12,1,2$ which is finite. Now we prove some results on Fibonacci numbers: First we claim $F_k+F_{k+4}=3F_{k+2}$. This is because \[F_k+F_{k+4}=F_k+F_{k+2}+F_{k+3}=F_k+F_{k+1}+2F_{k+2}=3F_{k+2}\]. Next we claim $F_{k+2}^2+1=F_kF_{k+4}$ for odd $k$. We prove this by induction: the base case $k=1$ works since $1\cdot 5=2^2+1$. Now \[-1=F_{k+2}^2-F_kF_{k+4}=F_{k+2}^2-(3F_{k+2}-F_{k+4})F_{k+4}=F_{k+4}^2-3F_{k+2}F_{k+4}+F_{k+2}^2=F_{k+4}^2-(3F_{k+4}-F_{k+2})F_{k+2}=F_{k+4}^2-F_{k+2}F_{k+6},\]completing the inductive step. Now we can check that for odd $k$ taking \[(a_1,a_2,a_3)=\left((F_k+F_{k+2})(F_{k+2}+F_{k+4}),\frac{F_k}{F_{k+2}}+1,\frac{F_{k+4}}{F_{k+2}}+1\right)\]works, which finishes.

I did 17SLN6 right before this