Example 12 (2003 Western China Mathematical Olympiad) Let $n$ be a given positive integer, find the smallest positive integer $u_{n}$, satisfying: for every positive integer $d$, the number of integers divisible by $d$ in any $u_{n}$ consecutive positive odd numbers is not less than the number of integers divisible by $d$ among the odd numbers $1,3,5, \cdots, 2 n-1$.
Lin
Pier
Ke
Mathematics
in
Number
Theory
Problem
The answer is $u_{n}=2 n-1$.
(1) First prove $u_{n} \geqslant 2 n-1$. Since $u_{1} \geqslant 1$, we may assume $n \geqslant 2$. In the sequence $1,3, \cdots, 2 n-1$, the number of terms divisible by $2 n-1$ is 1, and in the sequence $2(n+1)-1,2(n+2)-1, \cdots, 2(n+2 n-2)-1$, the number of terms divisible by $2 n-1$ is 0, so $u_{n} \geqslant 2 n-1$.
(2) Next prove $u_{n} \leqslant 2 n-1$. We only need to consider the case where $d$ is odd and $1 \leqslant d \leqslant 2 n-1$.
Consider the $2 n-1$ odd numbers: $2(a+1)-1,2(a+2)-1, \cdots, 2(a+2 n-1)-1$. Let $s$ and $t$ be integers such that
$$
\begin{array}{l}
(2 s-1) d \leqslant 2 n-1<(2 s+1) d, \\
(2 t-1) d<2(a+1)-1 \leqslant(2 t+1) d,
\end{array}
$$

Then the number of terms in $1,3, \cdots, 2 n-1$ that are divisible by $d$ is $s$, so we need to prove
$$
[2(t+s)-1] d \leqslant 2(a+2 n-1)-1
$$

to complete the proof. In fact, we have
$$
\begin{aligned}
(2(t+s)-1) d & =(2 t-1) d+(2 s-1) d+d \\
& \leqslant 2(a+1)-1+2 n-1+2 n-1 \\
& =2(a+2 n-1)-1,
\end{aligned}
$$

Therefore, $u_{n} \leqslant 2 n-1$.
In conclusion, we have $u_{n}=2 n-1$.
Note The key to this problem is to notice that in the sequence $2(n+1)-1,2(n+2)-1, \cdots, 2(n+2 n-2)-1$, the number of terms divisible by $2 n-1$ is 0, which is less than the number of terms in $1,3, \cdots, 2 n-1$ that are divisible by $2 n-1$. Thus, we get $u_{n} \geqslant 2 n-1$. Consequently, the problem is reduced to proving $u_{n} \leqslant 2 n-1$.