Alex and Zev are two members of a group of $2017$ friends who all know each other. Alex is trying to send a package to Zev. The delivery process goes as follows: Alex sends the package randomly to one of the people in the group. If this person is Zev, the delivery is done. Otherwise, the person who received the package also randomly sends it to someone in the group who hasn't held the package before and this process repeats until Zev gets the package. What is the expected number of deliveries made?
1. **Define the problem and the expected value**: We need to find the expected number of deliveries made until Zev receives the package. Let \( E_n \) be the expected number of deliveries for a group of \( n \) people.

2. **Initial conditions**: 
   - If there are only 2 people (Alex and Zev), the expected number of deliveries is 1 because Alex sends the package directly to Zev.
   - Therefore, \( E_2 = 1 \).

3. **Recursive relation**: 
   - For \( n \) people, Alex sends the package to one of the \( n-1 \) other people. The probability that the package is sent directly to Zev is \( \frac{1}{n-1} \).
   - If the package is not sent to Zev, it is sent to one of the remaining \( n-2 \) people, and the process continues. The expected number of deliveries in this case is \( 1 + E_{n-1} \) because one delivery has been made and we are left with \( n-1 \) people.

4. **Formulate the expected value**:
   - The expected number of deliveries can be written as:
     \[
     E_n = \frac{1}{n-1} \cdot 1 + \left(1 - \frac{1}{n-1}\right) \cdot (1 + E_{n-1})
     \]
   - Simplify the equation:
     \[
     E_n = \frac{1}{n-1} + \left(\frac{n-2}{n-1}\right) \cdot (1 + E_{n-1})
     \]
     \[
     E_n = \frac{1}{n-1} + \frac{n-2}{n-1} + \frac{n-2}{n-1} \cdot E_{n-1}
     \]
     \[
     E_n = 1 + \frac{n-2}{n-1} \cdot E_{n-1}
     \]

5. **Solve the recurrence relation**:
   - We know \( E_2 = 1 \).
   - For \( n = 3 \):
     \[
     E_3 = 1 + \frac{1}{2} \cdot E_2 = 1 + \frac{1}{2} \cdot 1 = 1 + \frac{1}{2} = \frac{3}{2}
     \]
   - For \( n = 4 \):
     \[
     E_4 = 1 + \frac{2}{3} \cdot E_3 = 1 + \frac{2}{3} \cdot \frac{3}{2} = 1 + 1 = 2
     \]
   - Continue this process to find a pattern or general formula.

6. **General formula**:
   - By induction or pattern recognition, we can see that:
     \[
     E_n = \sum_{k=1}^{n-1} \frac{1}{k}
     \]
   - This is the sum of the first \( n-1 \) terms of the harmonic series.

7. **Apply the formula to the given problem**:
   - For \( n = 2017 \):
     \[
     E_{2017} = \sum_{k=1}^{2016} \frac{1}{k}
     \]
   - The sum of the harmonic series \( \sum_{k=1}^{n} \frac{1}{k} \) is approximately \( \ln(n) + \gamma \), where \( \gamma \) is the Euler-Mascheroni constant. However, for large \( n \), this approximation is not necessary for exact values.

8. **Simplify the final expression**:
   - The sum \( \sum_{k=1}^{2016} \frac{1}{k} \) is not easily simplified without numerical methods, but it is known that the harmonic number \( H_{2016} \) is approximately \( \ln(2016) + \gamma \).

9. **Conclusion**:
   - The expected number of deliveries is:
     \[
     E_{2017} = \sum_{k=1}^{2016} \frac{1}{k} = H_{2016}
     \]
   - For large \( n \), this is approximately \( \ln(2016) + \gamma \).

The final answer is \( \boxed{H_{2016}} \).