Define a function on the positive integers recursively by $f(1) = 2$, $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$. What is $f(2017)$?

$\textbf{(A) } 2017 \qquad \textbf{(B) } 2018 \qquad \textbf{(C) } 4034 \qquad \textbf{(D) } 4035 \qquad \textbf{(E) } 4036$
1. Define the function \( f \) on the positive integers recursively:
   \[
   f(1) = 2
   \]
   \[
   f(n) = f(n-1) + 1 \quad \text{if } n \text{ is even}
   \]
   \[
   f(n) = f(n-2) + 2 \quad \text{if } n \text{ is odd and greater than 1}
   \]

2. Define a new function \( g(n) = f(n) - n \). We start with:
   \[
   g(1) = f(1) - 1 = 2 - 1 = 1
   \]

3. Analyze the behavior of \( g(n) \) for even \( n \):
   \[
   f(n) = f(n-1) + 1 \implies g(n) = f(n) - n = (f(n-1) + 1) - n = f(n-1) - (n-1) = g(n-1)
   \]
   Therefore, \( g(n) = g(n-1) \) when \( n \) is even.

4. Analyze the behavior of \( g(n) \) for odd \( n \):
   \[
   f(n) = f(n-2) + 2 \implies g(n) = f(n) - n = (f(n-2) + 2) - n = f(n-2) - (n-2) = g(n-2)
   \]
   Therefore, \( g(n) = g(n-2) \) when \( n \) is odd.

5. From the above steps, we see that \( g(n) \) is constant for all \( n \). Since \( g(1) = 1 \), we have:
   \[
   g(n) = 1 \quad \text{for all } n
   \]

6. Therefore, for any \( n \):
   \[
   f(n) = g(n) + n = 1 + n
   \]

7. Specifically, for \( n = 2017 \):
   \[
   f(2017) = 1 + 2017 = 2018
   \]

The final answer is \(\boxed{2018}\)