For real numbers $a, b$, and $c$ the polynomial $p(x) = 3x^7 - 291x^6 + ax^5 + bx^4 + cx^2 + 134x - 2$ has $7$ real roots whose sum is $97$. Find the sum of the reciprocals of those $7$ roots.
1. **Identify the polynomial and its roots:**
   The given polynomial is \( p(x) = 3x^7 - 291x^6 + ax^5 + bx^4 + cx^2 + 134x - 2 \). It has 7 real roots, say \( r_1, r_2, r_3, r_4, r_5, r_6, r_7 \).

2. **Sum of the roots:**
   By Vieta's formulas, the sum of the roots of the polynomial \( p(x) = 3x^7 - 291x^6 + ax^5 + bx^4 + cx^2 + 134x - 2 \) is given by the coefficient of \( x^6 \) divided by the leading coefficient (with a sign change). Therefore, the sum of the roots is:
   \[
   r_1 + r_2 + r_3 + r_4 + r_5 + r_6 + r_7 = \frac{291}{3} = 97
   \]
   This matches the given information that the sum of the roots is 97.

3. **Sum of the reciprocals of the roots:**
   The sum of the reciprocals of the roots of a polynomial \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \) is given by:
   \[
   \sum_{i=1}^n \frac{1}{r_i} = \frac{a_{n-1}}{a_n}
   \]
   Here, \( a_n = 3 \) (the coefficient of \( x^7 \)) and \( a_0 = -2 \) (the constant term). The product of the roots taken one at a time is given by:
   \[
   r_1 r_2 r_3 r_4 r_5 r_6 r_7 = \frac{-a_0}{a_n} = \frac{-(-2)}{3} = \frac{2}{3}
   \]

4. **Sum of the reciprocals:**
   The sum of the reciprocals of the roots is:
   \[
   \sum_{i=1}^7 \frac{1}{r_i} = \frac{\text{sum of the products of the roots taken 6 at a time}}{\text{product of all the roots}}
   \]
   By Vieta's formulas, the sum of the products of the roots taken 6 at a time is the coefficient of \( x \) (which is 134) divided by the leading coefficient (which is 3):
   \[
   \text{sum of the products of the roots taken 6 at a time} = \frac{134}{3}
   \]
   Therefore, the sum of the reciprocals of the roots is:
   \[
   \sum_{i=1}^7 \frac{1}{r_i} = \frac{\frac{134}{3}}{\frac{2}{3}} = \frac{134}{3} \times \frac{3}{2} = \frac{134}{2} = 67
   \]

The final answer is \(\boxed{67}\).