We define the ridiculous numbers recursively as follows:
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[*]1 is a ridiculous number.
[*]If $a$ is a ridiculous number, then $\sqrt{a}$ and $1+\sqrt{a}$ are also ridiculous numbers. 
[/list]
A closed interval $I$ is ``boring'' if
[list]
[*]$I$ contains no ridiculous numbers, and
[*]There exists an interval $[b,c]$ containing $I$ for which $b$ and $c$ are both ridiculous numbers.
[/list]
The smallest non-negative $l$ such that there does not exist a boring interval with length $l$ can be represented in the form $\dfrac{a + b\sqrt{c}}{d}$ where $a, b, c, d$ are integers, $\gcd(a, b, d) = 1$, and no integer square greater than 1 divides $c$. What is $a + b + c + d$?
1. **Identify the recursive definition of ridiculous numbers:**
   - \(1\) is a ridiculous number.
   - If \(a\) is a ridiculous number, then \(\sqrt{a}\) and \(1 + \sqrt{a}\) are also ridiculous numbers.

2. **Determine the least upper bound for ridiculous numbers:**
   - Start with \(1\). 
   - Apply the recursive rule: \(\sqrt{1} = 1\) and \(1 + \sqrt{1} = 2\).
   - Continue applying the rule to generate more ridiculous numbers:
     - \(\sqrt{2}\)
     - \(1 + \sqrt{2}\)
     - \(\sqrt{1 + \sqrt{2}}\)
     - \(1 + \sqrt{1 + \sqrt{2}}\)
     - And so on.

3. **Find the least upper bound for ridiculous numbers under 2:**
   - The sequence of ridiculous numbers generated from 1 includes \(\frac{1 + \sqrt{5}}{2}\), which is derived from the continued application of the recursive rule.
   - The least upper bound for ridiculous numbers under 2 is \(\frac{1 + \sqrt{5}}{2}\).

4. **Determine the smallest non-negative \(l\) such that there does not exist a boring interval with length \(l\):**
   - A boring interval \([b, c]\) must contain no ridiculous numbers and both \(b\) and \(c\) must be ridiculous numbers.
   - The smallest interval that can be formed without containing any ridiculous numbers is between two consecutive ridiculous numbers.
   - The smallest such interval is between \(\frac{1 + \sqrt{5}}{2}\) and the next ridiculous number, which is \(\frac{3 + \sqrt{5}}{2}\).

5. **Calculate the length \(l\) of this interval:**
   \[
   l = \frac{3 + \sqrt{5}}{2} - \frac{1 + \sqrt{5}}{2} = \frac{3 + \sqrt{5} - 1 - \sqrt{5}}{2} = \frac{2}{2} = 1
   \]
   - However, this interval does not satisfy the condition of being the smallest non-negative \(l\) such that there does not exist a boring interval with length \(l\).

6. **Re-evaluate the interval length:**
   - Consider the interval \(\left[\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}\right]\).
   - The length of this interval is:
   \[
   l = \frac{3 + \sqrt{5}}{2} - \frac{3 - \sqrt{5}}{2} = \frac{3 + \sqrt{5} - 3 + \sqrt{5}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}
   \]
   - This interval length is not the smallest non-negative \(l\).

7. **Find the correct interval length:**
   - The correct interval length is \(\frac{3 - \sqrt{5}}{2}\).

8. **Express \(l\) in the form \(\frac{a + b\sqrt{c}}{d}\):**
   - Here, \(a = 3\), \(b = -1\), \(c = 5\), and \(d = 2\).
   - Ensure \(\gcd(a, b, d) = 1\) and no integer square greater than 1 divides \(c\).

9. **Calculate \(a + b + c + d\):**
   \[
   a + b + c + d = 3 - 1 + 5 + 2 = 9
   \]

The final answer is \(\boxed{9}\)