Let $n$ be a given positive integer. Find the smallest positive integer $u_{n}$, satisfying: for every positive integer $d$, in any $u_{n}$ consecutive positive odd numbers, the number of numbers divisible by $d$ is not less than the number of numbers in $1,3,5, \cdots, 2 n-1$ that are divisible by $d$.
Take $d=2 n-1$. Among $1,3, \cdots, 2 n-1$, there is 1 number divisible by $d$, which is $2 n-1$. The $2 n-2$ consecutive positive odd numbers $2 n+1,2 n+3, \cdots, 2 n+2(2 n-1)-3$ are all greater than $2 n-1$ and less than $3(2 n-1)$. Therefore, none of them are multiples of $2 n-1$. So $u_{n} \geqslant 2 n-1$.
On the other hand, we prove that for any positive integer $d$, the number of multiples of $d$ in $2 n-1$ consecutive positive odd numbers is no less than the number of multiples of $d$ in $1,3, \cdots, 2 n-1$. When $d$ is even or $d>2 n-1$, there are no multiples of $d$ in $1,3, \cdots, 2 n-1$. The conclusion is obvious. Let $d$ be a positive odd number not greater than $2 n-1$. By the division algorithm, $2 n-1=q d+r, 0 \leqslant r<d$. (1) In $1,2,3, \cdots, 2 n-1$, there are exactly $q$ multiples of $d$, $d$, $2 d, 3 d, \cdots, q d$. (2) Therefore, in $1,3,5, \cdots, 2 n-1$ (3) there are at most $q$ multiples of $d$.
In any $d$ consecutive odd numbers $a, a+2, \cdots, a+2(d-1)$ (4), the difference between any two is $2 h, 0<h<d$. Since $d$ is odd, $2 h$ is not a multiple of $d$, so the $d$ numbers in (4) are distinct modulo $d$, and exactly one of them is a multiple of $d$. In $2 n-1$ consecutive positive odd numbers, every $d$ consecutive numbers form a group, and by (1), there are at least $q$ such groups, each containing exactly one multiple of $d$. Therefore, in $2 n-1$ consecutive positive odd numbers, there are at least $q$ multiples of $d$. This proves that $u_{n}=2 n-1$.