Find all natural numbers $ n$ such that the equation:

$ a_{n\plus{}1} x^2\minus{}2x \sqrt{a_1^2\plus{}a_2^2\plus{}...\plus{}a_{n\plus{}1}^2}\plus{}a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}0$ 

has real solutions for all real numbers $ a_1,a_2,...,a_{n\plus{}1}$.
To find all natural numbers \( n \) such that the equation:
\[ a_{n+1} x^2 - 2x \sqrt{a_1^2 + a_2^2 + \cdots + a_{n+1}^2} + a_1 + a_2 + \cdots + a_n = 0 \]
has real solutions for all real numbers \( a_1, a_2, \ldots, a_{n+1} \), we need to ensure that the discriminant of the quadratic equation is non-negative for all real \( a_i \).

1. **Identify the discriminant**:
   The general form of a quadratic equation \( ax^2 + bx + c = 0 \) has a discriminant given by \( \Delta = b^2 - 4ac \). For our equation:
   \[ a = a_{n+1}, \quad b = -2 \sqrt{a_1^2 + a_2^2 + \cdots + a_{n+1}^2}, \quad c = a_1 + a_2 + \cdots + a_n \]
   The discriminant is:
   \[
   \Delta = \left(-2 \sqrt{a_1^2 + a_2^2 + \cdots + a_{n+1}^2}\right)^2 - 4 a_{n+1} (a_1 + a_2 + \cdots + a_n)
   \]
   Simplifying, we get:
   \[
   \Delta = 4 (a_1^2 + a_2^2 + \cdots + a_{n+1}^2) - 4 a_{n+1} (a_1 + a_2 + \cdots + a_n)
   \]
   \[
   \Delta = 4 \left( (a_1^2 + a_2^2 + \cdots + a_{n+1}^2) - a_{n+1} (a_1 + a_2 + \cdots + a_n) \right)
   \]

2. **Ensure non-negativity of the discriminant**:
   We need:
   \[
   (a_1^2 + a_2^2 + \cdots + a_{n+1}^2) - a_{n+1} (a_1 + a_2 + \cdots + a_n) \geq 0
   \]
   Let \( S = a_1 + a_2 + \cdots + a_n \) and \( T = a_1^2 + a_2^2 + \cdots + a_{n+1}^2 \). Then the condition becomes:
   \[
   T - a_{n+1} S \geq 0
   \]

3. **Analyze the expression**:
   Consider the expression:
   \[
   P(a_1, a_2, \ldots, a_{n+1}) = n \left( (a_1^2 + a_2^2 + \cdots + a_{n+1}^2) - a_{n+1} (a_1 + a_2 + \cdots + a_n) \right)
   \]
   We can rewrite it as:
   \[
   P(a_1, a_2, \ldots, a_{n+1}) = \sum_{i=1}^{n} \left( a_{n+1} - \frac{n}{2} a_i \right)^2 + \left( n - \frac{n^2}{4} \right) \sum_{i=1}^{n} a_i^2
   \]

4. **Determine the condition for non-negativity**:
   For \( P(a_1, a_2, \ldots, a_{n+1}) \geq 0 \) for all real \( a_i \), we need:
   \[
   n - \frac{n^2}{4} \geq 0
   \]
   Solving this inequality:
   \[
   n - \frac{n^2}{4} \geq 0 \implies 4n - n^2 \geq 0 \implies n(4 - n) \geq 0
   \]
   This inequality holds for \( n = 1, 2, 3, 4 \).

5. **Check for \( n \geq 5 \)**:
   For \( n \geq 5 \), we have:
   \[
   n - \frac{n^2}{4} < 0
   \]
   Consider \( a_1 = a_2 = \cdots = a_n = 1 \) and \( a_{n+1} = \frac{n}{2} \):
   \[
   P(1, 1, \ldots, 1, \frac{n}{2}) = \sum_{i=1}^{n} \left( \frac{n}{2} - \frac{n}{2} \right)^2 + \left( n - \frac{n^2}{4} \right) n
   \]
   \[
   P(1, 1, \ldots, 1, \frac{n}{2}) = 0 + n \left( 1 - \frac{n}{4} \right) < 0
   \]
   Hence, the discriminant is negative for \( n \geq 5 \).

Conclusion:
The only values of \( n \) for which the equation has real solutions for all real \( a_1, a_2, \ldots, a_{n+1} \) are \( n = 1, 2, 3, 4 \).

The final answer is \( \boxed{ n = 1, 2, 3, 4 } \).