8. Let two strictly increasing sequences of positive integers $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy $a_{10}=b_{10}<2017$, for any positive integer $n$, there is $a_{n+2}=a_{n+1}+a_{n}, b_{n+1}=2 b_{n}$.
Then all possible values of $a_{1}+b_{1}$ are $\qquad$
8. 13, 20.

From the conditions, we know that \(a_{1}, a_{2}, b_{1}\) are all positive integers, and \(a_{1}b_{10} = 2^9 \times b_{1} = 512 b_{1}\), so \(b_{1} \in \{1, 2, 3\}\).
By repeatedly applying the recurrence relation of the sequence \(\{a_{n}\}\), we have
\[
\begin{array}{l}
a_{10} = a_{9} + a_{8} = 2 a_{8} + a_{7} = 3 a_{7} + 2 a_{6} \\
= 5 a_{6} + 3 a_{5} = 8 a_{5} + 5 a_{4} = 13 a_{4} + 8 a_{3} \\
= 21 a_{3} + 13 a_{2} = 34 a_{2} + 21 a_{1}.
\end{array}
\]

Thus, \(21 a_{1} \equiv a_{10} = b_{10} = 512 b_{1} \equiv 2 b_{1} \pmod{34}\).
Since \(13 \times 21 = 34 \times 8 + 1\), we have
\[
a_{1} \equiv 13 \times 21 a_{1} \equiv 13 \times 2 b_{1} = 26 b_{1} \pmod{34}.
\]

Noting that \(a_{1} < a_{2}\),
we have \(55 a_{1} < 34 a_{2} + 21 a_{1} = 512 b_{1}\)
\[
\Rightarrow a_{1} < \frac{512}{55} b_{1}.
\]

When \(b_{1} = 1\), equations (1) and (2) become
\[
a_{1} \equiv 26 \pmod{34}, \quad a_{1} < \frac{512}{55},
\]
which has no solution.
When \(b_{1} = 2\), equations (1) and (2) become
\[
a_{1} \equiv 52 \pmod{34}, \quad a_{1} < \frac{1024}{55},
\]
yielding the unique positive integer \(a_{1} = 18\), in which case \(a_{1} + b_{1} = 20\).
When \(b_{1} = 3\), equations (1) and (2) become
\[
a_{1} \equiv 78 \pmod{34}, \quad a_{1} < \frac{1536}{55},
\]
yielding the unique positive integer \(a_{1} = 10\), in which case \(a_{1} + b_{1} = 13\). In summary, the possible values of \(a_{1} + b_{1}\) are 13, 20.