China Western Mathematical Olympiad

(1) $u_n\ge 2n-1$. As $u_1=1$, we only need to consider $n\ge 2$. Since the number of integers divisible by $2n-1$ in $1,3,5,\dots ,2n-1$ is $1$ and that in $2(n+1)-1,2(n+2)-1,\dots ,2(n+2n-2)-1$ is $0$, then

$u_n\ge 2n-1$.

(2) $u_n\le 2n-1$. We only need to consider the case when $2\nmid d$ and $1\le d\le 2n-1$. For any $2n-1$ consecutive positive odd numbers; $2(a+1)-1,2(a+2)-1,\dots ,2(a+2n-1)-1$, let $s$ and $t$ be positive integers such that $$(2s-1)d\le 2n-1<(2s+1)d,$$ $$(2t-1)d\le 2(a+1)-1<(2t+1)d.$$ Then the number of integers divisible by $d$ in $1,3,5,\dots ,2n-1$ is $s$, and $$\begin{array}{rl}(2(t+s)-1)d& =(2t-1)d+(2s-1)d+d\\ & \le 2(a+1)-1+2n-1+2n-1\\ & =2(a+2n-1)-1.\end{array}$$ That means $u_n\le 2n-1$.