Team Selection Test

By inducting on $n$, we show that there exist relatively consistent $n$ distinct positive integers.

$=1+2$, and hence $2$ and $3$ is a consistent pair.

Now let $x_1<x_2<\cdots <x_n$ be relatively consistent $n$ distinct positive integers. Then for any given $i<j$ there exist distinct positive divisors of $x_i$, say $d_1,d_2,\dots ,d_k$, such that $d_1+d_2+\cdots +d_k=x_j$. Note that for any positive integer $a$, we have that $a{d}_1,a{d}_2,\dots ,a{d}_k$ are distinct divisors of $a{x}_i$ and their sum is equal to $a{x}_j$. Therefore, $a{x}_1<a{x}_2<\cdots <a{x}_n$ are relatively consistent for every positive integer $a$.

Let $b$ be an integer greater than $x_1$, and consider $b{x}_1$, $(b+1)x_1$, $(b+1)x_2,\dots ,(b+1)x_n$. By the observation above, $(b+1)x_1$, $(b+1)x_2,\dots ,(b+1)x_n$ are relatively consistent. Note that by the induction hypothesis, for any given $i>1$ there exist positive divisors of $x_1$, say $d_1,\dots ,d_m$, such that $d_1+d_2+\cdots +d_m=x_i$. Then, because of the choice of $b$, we see that $d_1,d_2,\dots ,d_m,b{d}_1,b{d}_2,\dots ,b{d}_m$ are distinct divisors of $b{x}_1$ and their sum is $(b+1)x_i$. Therefore, $b{x}_i$ and $(b+1)x_i$ are consistent for every $i>1$. Furthermore, $b{x}_1$ and $(b+1)x_1$ are consistent as well, since $x_1+b{x}_1=(b+1)x_1$.

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Alternative solution.

Let $a>1$ be an integer. It is easy to verify that $$a^{2017},a^{2017}+a^{2016},a^{2017}+a^{2016}+a^{2015},\dots ,a^{2017}+a^{2016}+\cdots +a+1$$ are pairwise relatively consistent.