For any real number $x$, let $\lfloor x \rfloor$ denote the integer part of $x$; $\{ x \}$ be the fractional part of $x$ ($\{x\}$ $=$ $x-$ $\lfloor x \rfloor$). Let $A$ denote the set of all real numbers $x$ satisfying 
$$\{x\} =\frac{x+\lfloor x \rfloor +\lfloor x + (1/2) \rfloor }{20}$$
If $S$ is the sume of all numbers in $A$, find $\lfloor S \rfloor$
To solve the problem, we need to find the set \( A \) of all real numbers \( x \) that satisfy the given equation:
\[ \{x\} = \frac{x + \lfloor x \rfloor + \lfloor x + \frac{1}{2} \rfloor}{20} \]

Let's denote \( x = I + f \), where \( I = \lfloor x \rfloor \) is the integer part and \( f = \{x\} \) is the fractional part. Therefore, the equation becomes:
\[ f = \frac{(I + f) + I + \lfloor I + f + \frac{1}{2} \rfloor}{20} \]

We need to consider two cases based on the value of \( f \):

### Case 1: \( 0 \leq f < \frac{1}{2} \)
In this case, \( \lfloor I + f + \frac{1}{2} \rfloor = I \). Therefore, the equation simplifies to:
\[ f = \frac{(I + f) + I + I}{20} \]
\[ f = \frac{2I + f}{20} \]
\[ 20f = 2I + f \]
\[ 19f = 2I \]
\[ f = \frac{2I}{19} \]

Since \( 0 \leq f < \frac{1}{2} \), we have:
\[ 0 \leq \frac{2I}{19} < \frac{1}{2} \]
\[ 0 \leq 2I < \frac{19}{2} \]
\[ 0 \leq I < \frac{19}{4} \]
\[ 0 \leq I < 4.75 \]

Thus, \( I \) can be 0, 1, 2, 3, or 4. The corresponding values of \( f \) are:
- \( I = 0 \): \( f = 0 \)
- \( I = 1 \): \( f = \frac{2}{19} \)
- \( I = 2 \): \( f = \frac{4}{19} \)
- \( I = 3 \): \( f = \frac{6}{19} \)
- \( I = 4 \): \( f = \frac{8}{19} \)

### Case 2: \( \frac{1}{2} \leq f < 1 \)
In this case, \( \lfloor I + f + \frac{1}{2} \rfloor = I + 1 \). Therefore, the equation simplifies to:
\[ f = \frac{(I + f) + I + (I + 1)}{20} \]
\[ f = \frac{3I + 1 + f}{20} \]
\[ 20f = 3I + 1 + f \]
\[ 19f = 3I + 1 \]
\[ f = \frac{3I + 1}{19} \]

Since \( \frac{1}{2} \leq f < 1 \), we have:
\[ \frac{1}{2} \leq \frac{3I + 1}{19} < 1 \]
\[ \frac{19}{2} \leq 3I + 1 < 19 \]
\[ \frac{17.5}{3} \leq I < 6 \]
\[ 5.8333 \leq I < 6 \]

Thus, \( I \) can only be 6. The corresponding value of \( f \) is:
- \( I = 6 \): \( f = \frac{19}{19} = 1 \)

However, \( f = 1 \) is not in the range \( \frac{1}{2} \leq f < 1 \), so there are no valid solutions in this case.

### Summing the Valid Solutions
The valid solutions for \( x \) are:
- \( x = 0 + 0 = 0 \)
- \( x = 1 + \frac{2}{19} = \frac{21}{19} \)
- \( x = 2 + \frac{4}{19} = \frac{42}{19} \)
- \( x = 3 + \frac{6}{19} = \frac{63}{19} \)
- \( x = 4 + \frac{8}{19} = \frac{84}{19} \)

The sum \( S \) of all these numbers is:
\[ S = 0 + \frac{21}{19} + \frac{42}{19} + \frac{63}{19} + \frac{84}{19} \]
\[ S = \frac{21 + 42 + 63 + 84}{19} \]
\[ S = \frac{210}{19} \]
\[ S = 11.0526 \]

Therefore, \( \lfloor S \rfloor = 11 \).

The final answer is \( \boxed{11} \)