Karl likes the number $17$ his favorite polynomials are monic quadratics with integer coefficients such that $17$ is a root of the quadratic and the roots differ by no more than $17$. Compute the sum of the coefficients of all of Karl's favorite polynomials. (A monic quadratic is a quadratic polynomial whose $x^2$ term has a coefficient of $1$.)
1. **Form of the Polynomial:**
   Since \(17\) is a root of the monic quadratic polynomial, we can write the polynomial as:
   \[
   P(x) = (x - 17)(x - r)
   \]
   Expanding this, we get:
   \[
   P(x) = x^2 - (17 + r)x + 17r
   \]
   Therefore, the polynomial is:
   \[
   P(x) = x^2 - (17 + r)x + 17r
   \]

2. **Condition on the Roots:**
   The roots differ by no more than \(17\). This means:
   \[
   |17 - r| \leq 17
   \]
   Solving this inequality, we get:
   \[
   -17 \leq 17 - r \leq 17
   \]
   This simplifies to:
   \[
   0 \leq r \leq 34
   \]
   Thus, \(r\) must be an integer between \(0\) and \(34\), inclusive.

3. **Sum of the Coefficients:**
   The sum of the coefficients of the polynomial \(P(x) = x^2 - (17 + r)x + 17r\) is:
   \[
   1 - (17 + r) + 17r = 1 - 17 - r + 17r = 1 - 17 + 16r = -16 + 16r
   \]
   We need to sum this expression for all integer values of \(r\) from \(0\) to \(34\):
   \[
   \sum_{r=0}^{34} (-16 + 16r)
   \]
   This can be split into two separate sums:
   \[
   \sum_{r=0}^{34} (-16) + \sum_{r=0}^{34} (16r)
   \]

4. **Calculating the Sums:**
   - The first sum is:
     \[
     \sum_{r=0}^{34} (-16) = -16 \times 35 = -560
     \]
   - The second sum is:
     \[
     \sum_{r=0}^{34} (16r) = 16 \sum_{r=0}^{34} r
     \]
     The sum of the first \(34\) integers is:
     \[
     \sum_{r=0}^{34} r = \frac{34 \times 35}{2} = 595
     \]
     Therefore:
     \[
     16 \sum_{r=0}^{34} r = 16 \times 595 = 9520
     \]

5. **Combining the Results:**
   Adding the two sums together, we get:
   \[
   -560 + 9520 = 8960
   \]

The final answer is \(\boxed{8960}\)