3 Let $n$ be a given positive integer, find the smallest positive integer $u_{n}$, satisfying: for any positive integer $d$, the number of integers divisible by $d$ in any $u_{n}$ consecutive odd numbers is not less than the number of integers divisible by $d$ in the odd numbers $1,3,5, \cdots, 2 n-1$. (Provided by Yonggao Chen)
The answer is $u_{n}=2 n \cdots 1$.
(1) First, prove that $u_{n} \geqslant 2 n-1$. By J $u_{1} \geqslant 1$. Therefore, assume $n \geq 2$. By J, in the sequence 1, 3, ..., $2 n-1$, the number of numbers 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 numbers divisible by $2 n-1$ is 0. Therefore, $u_{n} \geqslant 2 n-1$.
(2) Next, prove that $u_{n} \leqslant 2 n-1$. Only consider the case where $d$ is odd, $1 \leqslant d \leqslant 2 n-1$.

Consider the $2 n-1$ odd numbers: $2(a+1)-1, 2(a+2)-1, \cdots, 2(a+2 n-1)-1$. Let $s, t$ be integers such that
\[
\begin{array}{c}
(2 s-1) d \leqslant 2 n-1<(2 s+1) d, \\
(2 t-1) d<2(a+1)-1 \leqslant(2 t+1) d,
\end{array}
\]

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

In fact, we have
\[
\begin{aligned}
(2(t+s)-1) d & =(2 t-1) d+(2 s-1) d+d \\
& \leqslant 2(a+1)-1+2 n-1+2 n-1 \\
& =2(a+2 n-1)-1,
\end{aligned}
\]

Thus, $u_{n} \leqslant 2 n-1$.
In summary, we get $u_{n}=2 n-1$.