We claim the smallest possible common difference is $n!$. This is achievable with $a_i=i$ and $b_i=i+1$. Without loss of generality, let $\gcd(a_i,b_i)=1$. Otherwise, we can divide both $a_i$ and $b_i$ by their gcd. Next, let the arithmetic progression by $x, x+y, \dots, x+ny$. Then, by dividing adjacent terms, we get that $$\frac{a_i}{b_i} = \frac{x+(i-1)y}{x+iy}.$$However, since $\gcd(a_i,b_i)=1$, we get that $x+(i-1)y\mid a_i$, so in fact, $a_i\geq x+(i-1)y\geq i$. Finally, we have $$y=(b_1-a_1)\prod_{i=2}^na_i\geq 1\prod_{i=2}^n i=n!,$$as desired.

Let the common difference be $d_n$. We'll prove that $d_n \geq n!$ (equality occurs if $a_i=i$ and $b_i = i+1$) and $B_n = \prod\limits_{i=1}^{n} b_i \geq (n+1)!$ by induction. The base case $n=1$ is trivial and for the induction step from $n$ to $n+1$, note that the numbers $a_1a_2\ldots a_n$, $b_1a_2\ldots a_n$, $\ldots, b_1b_2\ldots b_n$ form an arithmetic progression because the numbers $a_1a_2\ldots a_na_{n+1}, b_1a_2\ldots a_na_{n+1}, \ldots, b_1b_2\ldots b_na_{n+1}$ do. Hence, by the induction hypothesis, \[d_{n+1} = d_na_{n+1} = B_n(b_{n+1}-a_{n+1}).\]However, $B_n$ is the $n$-th term in the arithmetic progression $a_1a_2\ldots a_n, b_1a_2\ldots a_n, \ldots, B_n$, so $B_n>nd_n$. From here, \[a_{n+1} = \frac{B_n}{d_n}(b_{n+1}-a_{n+1})>n(b_{n+1}-a_{n+1})\geq n.\]Therefore $d_{n+1} = d_na_{n+1} \geq d_n (n+1) \geq (n+1)!$ by the induction hypothesis. To conclude, $b_{n+1} > a_{n+1} \geq n+1$, so we have that $B_{n+1}\geq b_{n+1}(n+1)!\geq (n+2)!$ and the induction step is complete.

This problem was proposed by Matthew Brennan (SnowEverywhere), Canada.

We claim the answer is $n!$, achievable with $a_i = i, b_i = i+1$. Let $r_i = \frac {b_i}{a_i}$. We can deduce that \[(r_i - 1)r_{i-1} = r_{i-1} - 1 \implies r_i = \frac{2r_{i-1} - 1}{r_{i-1}}\]We let $a_1 = a, b_1 = b$ and get that \[r_i = \frac{ib - (i-1)a}{(i-1)b - (i-2)a}\text{ when } i \ge 2\]So the difference is at least \begin{align*} &\left(\frac{b}{a} - 1\right)(a)(2b-a)(3b-2a)\cdots ((n-1)b - (n-2)a) \\ &= (b-a)(2b - a) \cdots ((n-1)b - (n-2)a) \\ &\ge 1 \cdot 2 \cdots n = n! \end{align*}since $b-a \ge 1$ and $ib - (i-1)a \ge (i-1)(b-a) + b = i + 1$ since $b > a \ge 1$.

We see that the common difference is $$r=(b_i - a_i ) b_1b_2\dots b_{i-1} a_{i+1} \dots a_n $$ and we want smaller number can take it $r$ where $i$ is any positive integer and $n \ge i$ We see that if $$b_1b_2\dots b_{i-1}a_i\dots a_n,b_1b_2\dots b_{i}a_{i+1}\dots a_n ,b_1b_2\dots b_{i+1}a_{i+2}\dots a_n$$ are consecutive numbers in arithmetic sequence equal to $$ b_1b_2\dots b_{i-1}a_i\dots a_n + b_1b_2\dots b_{i+1}a_{i+2}\dots a_n = 2(b_1b_2\dots b_{i}a_{i+1}\dots a_n)$$ $$b_ib_{i+1}+a_ia_{i+1}=2(b_{i}a_{i+1})$$ $$\frac{a_i}{b_i} + \frac{b_{i+1}}{a_{i+1}}= 2 $$ We see that $gcd(a_i,b_i)=k$ where $k$ is any positive integer doesn't effect in this relation but we know that $b_i-a_i|r$ and we want smaller number $r$ so we can check that $gcd(a_i,b_i)=1$ where $i$ is a positive integer and $n \ge i$ from $b_ib_{i+1}+a_ia_{i+1}=2(b_{i}a_{i+1})$ we can see that $a_{i+1}|b_ib_{i+1} $ so $a_{i+1}|b_i$ and $b_i|a_ia_{i+1}$ so $ b_i|a_{i+1}$ so $b_i=a_{i+1}$ We get that $$a_{i+1}a_{i+2}+a_ia_{i+1}=2(a_{i+1})^2$$ $$a_{i+2}+a_i=2a_{i+1}, a_{i+2}-a_{i+1}=a_{i+1}-a_{i}=c$$ $$a_{i}=(n-1)c+a_{1}$$ $$r=(a_{i+1}-a_{i})a_2a_3\dots a_n=1(c+a_1)(2c+a_1)\dots ((n-1)c+a_1) \ge 1(2)(3)\dots (n)=n!$$ so $r\ge n!$ and if we let $$a_i=i,b_i=i+1$$ we see is a solution and $r=n!$ and we are done

We claim the answer is $n!$ which can be achieved by $a_i=i$ and $b_i=i+1$. We now show it is the smallest. Note that in order to be an arithmetic sequence: \[\frac{a_1a_2+b_1b_2}{2}=b_1a_2\]\[\frac{a_1a_2+b_1b_2}{2}=b_1a_2\]\[\dots\]\[\frac{a_{n-1}a_n+b_{n-1}b_n}{2}=b_{n-1}a_n.\]Thus: \[a_ia_{i+1}+b_ib_{i+1}=2b_ia_{i+1}\]\[\iff b_{i+1}-a_{i+1}=\frac{a_{i+1}}{b_i}(b_i-a_i).\]If we let $b_1-a_1=k$, then: \[b_1-a_1=k\]\[b_2-a_2=\frac{k}{b_1}\cdot a_2\]\[\dots\]\[b_n-a_n=\frac{k}{b_1}\cdot\frac{a_2}{b_2}\dots\frac{a_{n-1}}{b_{n-1}}\cdot a_n.\]As a result: \[D=k\cdot a_n\cdot a_{n-1}\dots a_3\cdot a_2.\] Let: \[f(i)=\frac{b_1}{k}\cdot\frac{b_2}{a_2}\dots\frac{b_{i}}{a_{i}}.\]Note: \[b_i-a_i=\frac{1}{f(i-1)}\cdot a_i\]\[\iff \frac{b_i}{a_i}-1=\frac{1}{f(i-1)}\]\[\iff f(i)=f(i-1)+1.\]Also: \[b_i-a_i\geq 1\]\[\implies a_i\geq f(i-1).\] Thus: \[f(1)=\frac{b_1}{k}>1\]\[\implies f(1)\geq 2\]\[\implies f(i)\geq i+1\]\[\implies a_i\geq f(i-1)\geq i\]\[\implies D\geq k\cdot n!\geq n!\]as desired $\blacksquare$

Note that scaling all the terms of an arithmetic progression doesn't change the fact that it is an arithmetic progression. With that in mind, scale the given arithmetic progression such that the first term is $1$, and the common difference $d=\frac{s}{t}$ for some $s,t \in \mathbb{N}$ such that $(s,t)=1$. Hence the arithmetic progression becomes, $$1, \frac{t+s}{t},\frac{t+2s}{t}, \cdots,\frac{t+ns}{t+(n-1)s}$$with $\frac{b_i}{a_i}=\frac{t+is}{t+(i-1)s}$ (Here, note that $(t+is,t+(i-1)s)=1$ because $(s,t)=1$) Now, note that all the terms of the arithmetic progression are integers. This forces that $a_i \ge t+(i-1)s$. Thus, $a_1 \cdots a_n \ge \prod_{i=1}^n (t+(i-1)s)$. In the equality case, the common difference becomes, $$\prod_{i=0}^{n-1}(s+it) \ge n!$$With equality when $s=t=1$

The answer is $n!$ which is achieved when $a_i=i$ and $b_i=i+1$ for all $1\leq i \leq n$. Choose positive integers $a$ and $d$ such that the arithmetic progression is $a,a+d,\dots, a+nd$. Let $g=\gcd(a,d)$, $a=ga'$, and $d=gd'$. Then by dividing consecutive terms, $\frac{b_i}{a_i}=\frac{a+id}{a+(i-1)d}$ thus $(a'+(i-1)d')|a_i$. Multiplying cyclically and dividing out by $a'$ gives that $$(a'+d)(a'+2d)\dots (a'+(n-1)d')|g$$Thus we have that $n!\leq g|d$, as desired.

By rewriting the definition for the original arithmetic progression we notice that $\{\frac{a_i}{b_i-a_i}, 1 \le i \le n\}$ is an arithmetic progression with difference 1. Hence, for each $1 \le i \le n$, $a_i$ > $(b_i-a_i)(i-1)$, which implies $a_i \ge i$. Then the difference is $(b_1-a_1)a_2a_3 \ldots a_n \ge 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n = n!$

The answer is $\boxed{n!}$, with construction $a_k=k$ and $b_k=k+1$. Clearly this works. Let $x$ be the first term, and let $y$ be the common difference. Let $d=\gcd(x,y)$, and let $x=ds$, $y=dt$. Now note that \[\frac{b_k}{a_k}=\frac{x+ky}{x+(k-1)y}=\frac{s+kt}{s+(k-1)t}.\]Note that this is a reduced fraction because $\gcd(s+kt,s+(k-1)t)=\gcd(s+kt,t)=\gcd(s,t)=1$. Therefore, \[x=a_1a_2\dots a_n\ge (s)(s+t)(s+2t)\dots (s+(n-1)t),\]so \[d\ge (s+t)(s+2t)\dots (s+(n-1)t)\ge (1+1)(1+2)\dots (1+(n-1))=n!,\]so $y=dt\ge d\ge n!$. $\blacksquare$

The answer is $n!$ which is achieved for $a_i=i$ and $b_i=i+1$. First of all, it's easy to see that $b_i\ge a_i+1$ for all $i$. Notice that since the sequence is an arithmetic progression, the term with $k$ $b$'s plus the term with $k+2$ $b$'s is equal to twice the term with $k+1$ $b$'s. After dividing by the common terms this yields \[a_ka_{k+1}+b_kb_{k+1}=2b_ka_{k+1}\iff \frac{b_k}{b_k-a_k}=\frac{a_{k+1}}{b_{k+1}-a_{k+1}}\]Now let $c_k=\frac{a_k}{b_k-a_k}$. We have that \[c_k=c_1+k-1, \ \forall k\]And since $c_1>0$ We get $\frac{a_k}{b_k-a_k}> k-1$ and this implies that $a_k>k-1$, or $a_k\ge k$. Finally, note that the common difference is \[(b_1-a_1)a_2\ldots a_n\ge n!\]And this completes the proof. $\blacksquare$

MarkBcc168 wrote: We claim the smallest possible common difference is $n!$. This is achievable with $a_i=i$ and $b_i=i+1$. Without loss of generality, let $\gcd(a_i,b_i)=1$. Otherwise, we can divide both $a_i$ and $b_i$ by their gcd. Next, let the arithmetic progression by $x, x+y, \dots, x+ny$. Then, by dividing adjacent terms, we get that $$\frac{a_i}{b_i} = \frac{x+(i-1)y}{x+iy}.$$However, since $\gcd(a_i,b_i)=1$, we get that $x+(i-1)y\mid a_i$, so in fact, $a_i\geq x+(i-1)y\geq i$. Finally, we have $$y=(b_1-a_1)\prod_{i=2}^na_i\geq 1\prod_{i=2}^n i=n!,$$as desired. Sorry but, shouldn't it be $a_i | x+(i-1)y$?

The smallest possible difference is $n!$. We will show the bound and give a construction simultaneously. Let $d$ be the common difference. Notice that \[a_2a_3 \cdots a_n(b_1-a_1) = d = b_1a_3\cdots a_n(b_2-a_2)\]so $\frac{b_1-a_1}{b_2 - a_2} = \frac{b_1}{a_2}$. Similar equations hold for each index $i$. Letting $c_i = b_i - a_i$ for each $1 \leq i \leq n$ (noting that $c_i > 0$), the equations rewrite as $\frac{c_i}{c_i+a_i} = \frac{c_{i+1}}{a_{i+1}}$ for each $i$. Now let $d_i = \gcd(a_i, c_i)$ and $a_i = d_ix_i$, $c_i = d_i z_i$ for each $i$, with $x_i$ and $z_i$ relatively prime. Then $\frac{z_i}{x_i+z_i} = \frac{z_{i+1}}{x_{i+1}}$ for each $i$, which implies that $z_i = z_{i+1} = z$ for some $z$ and $x_i = x_i + z_i = x_1 + (i-1)z$. In particular, \[d = a_2a_3\cdots a_n c_1 = d_1d_2\cdots d_n \prod_{i=2}^n (x_1 + (i-1)z).\]The values $x_1 + (i-1)z$ are positive and distinct for each $i$, hence they multiply to at least $n!$. Equality holds when $x_1 = 0$ and $z = 1$, with all the $d_i$ equal to $1$.