Solution:

Let $a_1 = a$ be the first term of such arithmetic sequence and $d > 0$ be its common difference. Then the condition $a_{a_{20}} = 17$ is equivalent to
$$a_{a_{20}} = a + (a_{20} - 1)d = a + (a + 19d - 1)d = a(1 + d) + 19d^2 - d.$$
Solving for $a$, we get
$$a = \frac{-19d^2 + d + 17}{d + 1} = -19d + 20 - \frac{3}{d + 1}.$$
Since $a$ and $d$ are integers, $d + 1$ must divide $3$. With $d > 0$, this forces $d + 1 = 3$ or $d = 2$, so
$$a = -19(2) + 20 - 1 = -19.$$
Hence,
$$a_{2017} = a + 2016d = -19 + 2016 \times 2 = 4013.$$