A [u]positive sequence[/u] is a finite sequence of positive integers. [u]Sum of a sequence[/u] is the sum of all the elements in the sequence. We say that a sequence $A$ can be [u]embedded[/u] into another sequence $B$, if there exists a strictly increasing function $$\phi : \{1,2, \ldots, |A|\} \rightarrow 
\{1,2, \ldots, |B|\},$$ such that $\forall i \in \{1, 2, \ldots ,|A|\}$, $$A[i] \leq B[\phi(i)],$$ where $|S|$ denotes the length of
a sequence $S$. For example, $(1,1,2)$ can be embedded in $(1,2,3)$, but $(3,2,1)$ can not be in $(1,2,3)$\\
Given a positive integer $n$, construct a positive sequence $U$ with sum $O(n \, \log \, n)$, such that all the positive sequences with sum $n$, can be embedded into $U$.\\
1. **Define the Sequence Construction:**
   The sequence \( U \) can be constructed recursively. Let the dot ( \(\cdot\) ) denote the concatenation of sequences. The recursive definition is given by:
   \[
   Seq(n) = Seq(\lfloor n/2 \rfloor) \cdot n \cdot Seq(\lfloor n/2 \rfloor)
   \]
   This means that for any positive integer \( n \), the sequence \( Seq(n) \) is formed by concatenating \( Seq(\lfloor n/2 \rfloor) \), the integer \( n \), and another \( Seq(\lfloor n/2 \rfloor) \).

2. **Embedding a Sequence \( A \) into \( Seq(n) \):**
   To embed a sequence \( A \) with sum \( n \) into \( Seq(n) \), we need to find an index \( i \) such that:
   \[
   A[1] + A[2] + \ldots + A[i] \leq \lfloor n/2 \rfloor
   \]
   and
   \[
   A[1] + A[2] + \ldots + A[i] + A[i+1] > \lfloor n/2 \rfloor
   \]
   This ensures that the sum of the first \( i \) elements of \( A \) can be embedded into the first part of \( Seq(n) \), the element \( A[i+1] \) can be embedded into the middle part (which is \( n \)), and the remaining elements can be embedded into the last part of \( Seq(n) \).

3. **Sum of the Sequence \( U \):**
   By the standard mergesort recursion, the sum of the sequence \( U \) is \( O(n \log n) \). This can be seen by analyzing the recursive construction:
   \[
   T(n) = 2T(\lfloor n/2 \rfloor) + n
   \]
   Using the Master Theorem for divide-and-conquer recurrences, we find that \( T(n) = O(n \log n) \).

4. **Explicit Construction for \( n = 2^k \):**
   For \( n \) of the form \( 2^k \), the sequence can be explicitly constructed by:
   \[
   a_i = 2^{v_2(i)}
   \]
   where \( v_2(i) \) denotes the highest power of 2 that divides \( i \). This construction ensures that the sequence \( U \) has the desired properties and sum \( O(n \log n) \).

The final answer is \( \boxed{ Seq(n) = Seq(\lfloor n/2 \rfloor) \cdot n \cdot Seq(\lfloor n/2 \rfloor) } \)