Find all real number $x$ which could be represented as
$x = \frac{a_0}{a_1a_2 . . . a_n} + \frac{a_1}{a_2a_3 . . . a_n} + \frac{a_2}{a_3a_4 . . . a_n} + . . . + \frac{a_{n-2}}{a_{n-1}a_n} + \frac{a_{n-1}}{a_n}$ , with $n, a_1, a_2, . . . . , a_n$ are positive integers and $1 = a_0 \leq a_1 < a_2 < . . .  < a_n$
To find all real numbers \( x \) that can be represented as 
\[ x = \frac{a_0}{a_1a_2 \cdots a_n} + \frac{a_1}{a_2a_3 \cdots a_n} + \frac{a_2}{a_3a_4 \cdots a_n} + \cdots + \frac{a_{n-2}}{a_{n-1}a_n} + \frac{a_{n-1}}{a_n}, \]
where \( n, a_1, a_2, \ldots, a_n \) are positive integers and \( 1 = a_0 \leq a_1 < a_2 < \cdots < a_n \), we proceed as follows:

1. **Representation as Continued Fraction:**
   Let us write \( x \) in the form of a continued fraction:
   \[ x = [1; a_1, a_2, \ldots, a_n]. \]
   This notation means that \( x \) is represented as:
   \[ x = \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \cdots + \frac{1}{a_n}}}}. \]

2. **Range of \( x \):**
   Since \( a_1, a_2, \ldots, a_n \) are positive integers and \( 1 = a_0 \leq a_1 < a_2 < \cdots < a_n \), it is clear that \( x \) is a positive rational number. We need to determine the range of \( x \).

3. **Upper Bound Analysis:**
   Note that for any positive integer \( a \), we have:
   \[ [1; a] = \frac{1}{a} \leq 1. \]
   Therefore, for \( n \geq 2 \), if \( [1; a_1, \ldots, a_n] > 1 \), then:
   \[ [1; a_1, \ldots, a_{n-1}] = a_n [1; a_1, \ldots, a_n] - a_{n-1} > a_n - a_{n-1} \geq 1. \]
   This implies that \( [1; a_1, \ldots, a_n] > 1 \) is impossible. Hence, all these numbers are \( \leq 1 \).

4. **Lower Bound Analysis:**
   Note that:
   \[ \frac{1}{q} = [1; q]. \]
   Also, for any \( q \geq p > 1 \), we have:
   \[ \frac{p}{q} = [1; 1, 2, 3, \ldots, p-2, p-1, q]. \]
   This shows that any rational number in the interval \( (0, 1] \) can be represented in the given form.

5. **Conclusion:**
   Therefore, the required set of numbers \( x \) is exactly the set of all rational numbers in the interval \( (0, 1] \).

The final answer is \( \boxed{ (0, 1] } \).