Let $$C=\{1,22,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$$ and let
$$S=\{4,5,9,14,23,37\}$$ Find two sets $A$ and $B$ with the properties
(a) $A\cap B=\emptyset$.
(b) $A\cup B=C$.
(c) The sum of two distinct elements of $A$ is not in $S$.
(d) The sum of two distinct elements of $B$ is not in $S$.

To solve this problem, we need to find two sets \( A \) and \( B \) such that:

1. \( A \cap B = \emptyset \)
2. \( A \cup B = C \)
3. The sum of any two distinct elements of \( A \) is not in \( S \)
4. The sum of any two distinct elements of \( B \) is not in \( S \)

Given:
\[ C = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\} \]
\[ S = \{4, 5, 9, 14, 23, 37\} \]

Let's verify the proposed solution:
\[ A = \{1, 2, 5, 6, 10, 11, 14, 15, 16, 19, 20\} \]
\[ B = C \setminus A = \{3, 4, 7, 8, 9, 12, 13, 17, 18\} \]

1. **Check \( A \cap B = \emptyset \)**:
   \[ A \cap B = \{1, 2, 5, 6, 10, 11, 14, 15, 16, 19, 20\} \cap \{3, 4, 7, 8, 9, 12, 13, 17, 18\} = \emptyset \]
   This condition is satisfied.

2. **Check \( A \cup B = C \)**:
   \[ A \cup B = \{1, 2, 5, 6, 10, 11, 14, 15, 16, 19, 20\} \cup \{3, 4, 7, 8, 9, 12, 13, 17, 18\} = C \]
   This condition is satisfied.

3. **Check the sum of any two distinct elements of \( A \) is not in \( S \)**:
   - Possible sums of two distinct elements in \( A \):
     \[ 1+2=3, \; 1+5=6, \; 1+6=7, \; 1+10=11, \; 1+11=12, \; 1+14=15, \; 1+15=16, \; 1+16=17, \; 1+19=20, \; 1+20=21 \]
     \[ 2+5=7, \; 2+6=8, \; 2+10=12, \; 2+11=13, \; 2+14=16, \; 2+15=17, \; 2+16=18, \; 2+19=21, \; 2+20=22 \]
     \[ 5+6=11, \; 5+10=15, \; 5+11=16, \; 5+14=19, \; 5+15=20, \; 5+16=21, \; 5+19=24, \; 5+20=25 \]
     \[ 6+10=16, \; 6+11=17, \; 6+14=20, \; 6+15=21, \; 6+16=22, \; 6+19=25, \; 6+20=26 \]
     \[ 10+11=21, \; 10+14=24, \; 10+15=25, \; 10+16=26, \; 10+19=29, \; 10+20=30 \]
     \[ 11+14=25, \; 11+15=26, \; 11+16=27, \; 11+19=30, \; 11+20=31 \]
     \[ 14+15=29, \; 14+16=30, \; 14+19=33, \; 14+20=34 \]
     \[ 15+16=31, \; 15+19=34, \; 15+20=35 \]
     \[ 16+19=35, \; 16+20=36 \]
     \[ 19+20=39 \]
   - None of these sums are in \( S \). This condition is satisfied.

4. **Check the sum of any two distinct elements of \( B \) is not in \( S \)**:
   - Possible sums of two distinct elements in \( B \):
     \[ 3+4=7, \; 3+7=10, \; 3+8=11, \; 3+9=12, \; 3+12=15, \; 3+13=16, \; 3+17=20, \; 3+18=21 \]
     \[ 4+7=11, \; 4+8=12, \; 4+9=13, \; 4+12=16, \; 4+13=17, \; 4+17=21, \; 4+18=22 \]
     \[ 7+8=15, \; 7+9=16, \; 7+12=19, \; 7+13=20, \; 7+17=24, \; 7+18=25 \]
     \[ 8+9=17, \; 8+12=20, \; 8+13=21, \; 8+17=25, \; 8+18=26 \]
     \[ 9+12=21, \; 9+13=22, \; 9+17=26, \; 9+18=27 \]
     \[ 12+13=25, \; 12+17=29, \; 12+18=30 \]
     \[ 13+17=30, \; 13+18=31 \]
     \[ 17+18=35 \]
   - None of these sums are in \( S \). This condition is satisfied.

Since all conditions are satisfied, the sets \( A \) and \( B \) are valid.

The final answer is \( \boxed{ A = \{1, 2, 5, 6, 10, 11, 14, 15, 16, 19, 20\} } \) and \( B = \{3, 4, 7, 8, 9, 12, 13, 17, 18\} \)