3. 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 $d$ in the odd numbers $1,3,5, \cdots$, $2 n-1$.
3. $u_{n}=2 n-1$.
(1) First, prove $u_{n} \geqslant 2 n-1$.

Since $u_{1} \geqslant 1$, assume $n \geqslant 2$. In the sequence $1,3, \cdots, 2 n-1$, the number of terms divisible by $2 n-1$ is 1. 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. Therefore, $u_{n} \geqslant 2 n-1$.
(2) Next, prove $u_{n} \leqslant 2 n-1$.
Consider $d$ as an odd number 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
$$
\begin{array}{l}
(2 s-1) d \leqslant 2 n-1<(2 s+1) d, \\
(2 t-1) d<2(a+1)-1 \leqslant(2 t+1) d .
\end{array}
$$

Thus, the number of terms in $1,3, \cdots, 2 n-1$ that are divisible by $d$ is $s$. Therefore, it suffices to prove
$$
[2(t+s)-1] d \leqslant 2(a+2 n-1)-1
$$

Indeed, we have
$$
\begin{array}{l}
{[2(t+s)-1] d=(2 t-1) d+(2 s-1) d+d} \\
\leqslant 2(a+1)-3+2 n-1+2 n-1 \\
=2(a+2 n-1)-1 .
\end{array}
$$

Therefore, $u_{n} \leqslant 2 n-1$.