Let $n$ be a fixed positive integer. 
- Show that there exist real polynomials $p_1, p_2, p_3, \cdots, p_k \in \mathbb{R}[x_1, \cdots, x_n]$ such that 
        \[(x_1 + x_2 + \cdots + x_n)^2 + p_1(x_1, \cdots, x_n)^2 + p_2(x_1, \cdots, x_n)^2 + \cdots + p_k(x_1, \cdots, x_n)^2 = n(x_1^2 + x_2^2 + \cdots + x_n^2)\]
- Find the least natural number $k$, depending on $n$, such that the above polynomials $p_1, p_2, \cdots, p_k$ exist.  
1. **Initial Setup and Assumptions:**
   We need to show that there exist real polynomials \( p_1, p_2, \ldots, p_k \in \mathbb{R}[x_1, \ldots, x_n] \) such that 
   \[
   (x_1 + x_2 + \cdots + x_n)^2 + p_1(x_1, \cdots, x_n)^2 + p_2(x_1, \cdots, x_n)^2 + \cdots + p_k(x_1, \cdots, x_n)^2 = n(x_1^2 + x_2^2 + \cdots + x_n^2).
   \]
   We aim to find the least natural number \( k \) for which this is possible.

2. **Degree of Polynomials:**
   First, observe that all \( p_j \) must have degree 1. Let \( p_j(x) = a_{1j}x_1 + a_{2j}x_2 + \cdots + a_{nj}x_n \).

3. **Vector Representation:**
   Define vectors \( v_i = (1, a_{i1}, a_{i2}, \ldots, a_{ik}) \) and normalize them to get \( u_i = \frac{v_i}{\sqrt{n}} \).

4. **Conditions:**
   The given condition is equivalent to:
   - \( u_i \cdot u_j = 0 \) for \( i \neq j \) (orthogonality),
   - \( |u_i| = 1 \) for all \( i \) (unit vectors),
   - \( \sum_{t=1}^n u_i = (\sqrt{n}, 0, \ldots, 0) \),
   - The first element of each \( u_i \) is \( \frac{1}{\sqrt{n}} \).

5. **Dimensionality and Linear Independence:**
   The problem reduces to finding \( n \) linearly independent vectors in \( k+1 \) dimensions. Hence, \( k \geq n-1 \).

6. **Inductive Construction:**
   - For \( n = 2 \), the vectors \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) and \( \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \) work.
   - Assume the construction works for \( n-1 \).
   - For \( n \) dimensions, split the vector \( (\sqrt{n}, 0, \ldots, 0) \) into two perpendicular components: one of length 1 and the other of length \( \sqrt{n-1} \). Call the former \( u_1 \) and the latter \( w \).
   - Apply the assumption for \( n-1 \) in the \( n-1 \) dimensional space perpendicular to \( u_1 \) with \( w \) as \( (\sqrt{n-1}, 0, \ldots, 0) \) in that space.

7. **Conclusion:**
   By induction, we can construct the required vectors, proving that \( k = n-1 \) is the least natural number for which the polynomials \( p_1, p_2, \ldots, p_k \) exist.

\(\blacksquare\)

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