$6$ cats can eat $6$ fish in $1$ day, and $c$ cats can eat $91$ fish in $d$ days. Given that $c$ and $d$ are both whole numbers, and the number of cats, $c$, is more than $1$ but less than $10$, find $c + d$.
1. We start by understanding the rate at which the cats eat the fish. From the first statement, we know that 6 cats can eat 6 fish in 1 day. Therefore, the rate of one cat eating fish is:
   \[
   \text{Rate per cat} = \frac{6 \text{ fish}}{6 \text{ cats} \times 1 \text{ day}} = 1 \text{ fish per cat per day}
   \]

2. From the second statement, we need to find the number of cats \( c \) and the number of days \( d \) such that \( c \) cats can eat 91 fish in \( d \) days. Using the rate we found, we can set up the equation:
   \[
   \text{(rate per cat)} \times \text{(number of cats)} \times \text{(number of days)} = \text{(total fish)}
   \]
   Substituting the known values, we get:
   \[
   1 \times c \times d = 91 \implies cd = 91
   \]

3. Next, we need to find the integer pairs \((c, d)\) that satisfy \( cd = 91 \) and the condition \( 1 < c < 10 \). We perform the prime factorization of 91:
   \[
   91 = 7 \times 13
   \]

4. Considering the factor pairs of 91, we have:
   \[
   (c, d) = (7, 13) \quad \text{or} \quad (c, d) = (13, 7)
   \]
   Since \( c \) must be between 1 and 10, the only valid pair is:
   \[
   (c, d) = (7, 13)
   \]

5. Finally, we calculate \( c + d \):
   \[
   c + d = 7 + 13 = 20
   \]

The final answer is \(\boxed{20}\).