Let $ n$ be a given positive integer.  Find the smallest positive integer $ u_n$ such that for any positive integer $ d$, in any $ u_n$ consecutive odd positive integers, the number of them that can be divided by $ d$ is not smaller than the number of odd integers among $ 1, 3, 5, \ldots, 2n \minus{} 1$ that can be divided by $ d$.
1. **Restate the problem in a more structured way:**

   Let \( n \) be a positive integer. Define the set \( N_{m,k} \), where \( m \) is an odd positive integer, to be the set of the \( k \) smallest odd integers that are at least \( m \). For an arbitrary positive integer \( d \), denote by \( N_{m,k}(d) \) the number of integers in the set \( N_{m,k} \) that are divisible by \( d \). Find the smallest \( u_n \in \mathbb{N} \) (in terms of \( n \)) such that \( N_{m,u_n}(d) \ge N_{1,n}(d) \) for every positive integer \( d \) and odd positive integer \( m \).

2. **Identify the pattern and conjecture:**

   The first few values for \( u_n \) are \( 1, 3, 5, 7, 9 \), so we conjecture \( u_n = 2n - 1 \).

3. **Verify the conjecture:**

   - Clearly, \( N_{1,n}(2n-1) = 1 \) because of the presence of \( 2n-1 \) in the set \( N_{1,n} \). 
   - The next odd multiple is \( 3(2n-1) = 2(3n-1) - 1 \), which means that \( N_{2n+1,2n-2}(2n-1) = 0 \). 
   - Therefore, \( u_n \ge 2n-1 \).

4. **Show that \( u_n = 2n-1 \) works:**

   - Note that \( N_{1,n}(d) = 0 \) for \( d \not\in \{1, 3, 5, \ldots, 2n-1\} \), so we only need to worry about \( d \) in this set.
   - Let \( m \) be an arbitrary odd positive integer. Then the set \( N_{m,2n-1} \) consists of the integers \( m, m+2, m+4, \ldots, m+2(2n-1)-2 \).

5. **Prove that one of the first \( d \) of these integers is divisible by \( d \):**

   - Suppose none of the first \( d \) integers are multiples of \( d \). Then none are equal to \( 0 \pmod{d} \), and by the pigeonhole principle, there are two equal \( \pmod{d} \), say \( m+a \) and \( m+b \) where \( 0 \le a < b \le 2d-2 \).
   - This means \( d \mid (b-a) \). Since \( 2d-2 \ge b-a > 0 \), the only multiple of \( d \) that \( b-a \) can possibly be is \( d \) itself.
   - Now \( a \) and \( b \) are both even, so \( b-a \) is even, meaning \( d \) is even, which is a contradiction since we assumed \( d \in \{1, 3, 5, \ldots, 2n-1\} \).

6. **Conclude that one of the first \( d \) of these integers is divisible by \( d \):**

   - Let \( m+a \) be the smallest integer in the set \( N_{m,2n-1} \) such that \( d \mid (m+a) \). We have just proven \( a \le 2d-2 \).

7. **Determine the number of integers in \( N_{m,2n-1} \) divisible by \( d \):**

   - Let \( h \) be the largest integer such that \( m+a+2hd \le m+4n-4 \). Then the integers in \( N_{m,2n-1} \) which are divisible by \( d \) are \( m+a, m+a+2d, m+a+4d, \ldots, m+a+2hd \).
   - It follows that \( N_{m,2n-1}(d) = h+1 \).

8. **Compare with \( N_{1,n}(d) \):**

   - Let \( b \) be the largest odd integer such that \( bd \le 2n-1 \). If \( b = 2c-1 \), then it follows that \( N_{1,n}(d) = c \ge 1 \).
   - Now \( (2c-1)d \le 2n-1 \) so \( 2(2c-1)d \le 4n-2 \).
   - Since \( h \) is the largest integer such that \( m+a+2hd \le m+4n-4 \), we have \( a+2(h+1)d \ge 4n-4 \).
   - Now \( 2d-2 \ge a \) meaning \( 2d-2+2(h+1)d \ge a+2(h+1)d \ge 4n-4 \ge 2(2c-1)d-2 \).
   - Then it follows that \( h+1 \ge 2c-2 \ge c \) for \( c > 1 \). When \( c = 1 \), we have \( d = 2n-1 \) in which case \( h = 0 \) so \( h+1 = 1 \ge 1 = c \).

9. **Conclude that \( N_{m,2n-1}(d) \ge N_{1,n}(d) \):**

   - Therefore, \( u_n = 2n-1 \) works.

The final answer is \( \boxed{ u_n = 2n-1 } \).