Consider the set $E = \{5, 6, 7, 8, 9\}$. For any partition ${A, B}$ of $E$, with both $A$ and $B$ non-empty, consider the number obtained by adding the product of elements of $A$ to the product of elements of $B$. Let $N$ be the largest prime number amonh these numbers. Find the sum of the digits of $N$.
1. **Identify the constraints for \( N \) to be a prime number:**
   - Since \( N \) must be a prime number, it cannot be even. Therefore, \( 6 \) and \( 8 \) cannot belong to different sets because their product would be even, making \( N \) even.
   - Similarly, \( 6 \) and \( 9 \) must belong to the same set because their product would be divisible by \( 3 \), making \( N \) divisible by \( 3 \).

2. **Determine the set memberships:**
   - From the above constraints, \( 6 \), \( 8 \), and \( 9 \) must belong to the same set to avoid making \( N \) even or divisible by \( 3 \).

3. **Consider the possible partitions:**
   - Without loss of generality, assume \( 5 \in A \). We then have two cases to consider:
     - Case 1: \( A = \{5, 6, 8, 9\} \) and \( B = \{7\} \)
     - Case 2: \( A = \{5, 7\} \) and \( B = \{6, 8, 9\} \)

4. **Calculate \( N \) for each case:**
   - **Case 1:**
     \[
     N = 5 \cdot 6 \cdot 8 \cdot 9 + 7 = 2160 + 7 = 2167
     \]
     - Check if \( 2167 \) is a prime number:
       \[
       2167 = 11 \cdot 197
       \]
       Since \( 2167 \) is not a prime number, this case is invalid.

   - **Case 2:**
     \[
     N = 5 \cdot 7 + 6 \cdot 8 \cdot 9 = 35 + 432 = 467
     \]
     - Check if \( 467 \) is a prime number:
       - \( 467 \) is not divisible by \( 2, 3, 5, 7, 11, 13, 17, 19 \) (all primes less than \( \sqrt{467} \approx 21.6 \)).
       - Therefore, \( 467 \) is a prime number.

5. **Sum the digits of \( N \):**
   \[
   4 + 6 + 7 = 17
   \]

The final answer is \( \boxed{17} \)