Solution:
Let us prove by induction that for any positive integer $k$, there is an $N$ for which
$$\sum_{n=1}^{N} a_{n} \geq k a_{1}$$
The base case, $k=1$ and $N=1$, is trivial. Given that (1) is true for a given $k$, we have
$$k a_{1} \leq \sum_{n=1}^{N} a_{n} \leq \sum_{n=1}^{N}\left(a_{2 n}+a_{2 n+1}\right)=\sum_{\substack{2 \leq n \leq 2 N \\ n \text{ even }}} a_{n}+\sum_{\substack{3 \leq n \leq 2 N+1 \\ n \text{ odd }}} a_{n}=\sum_{n=2}^{2 N+1} a_{n}$$
Adding $a_{1}$, we get
$$\sum_{n=1}^{2 N+1} a_{n} \geq(k+1) a_{1}$$
as desired.

The problem now follows by taking $k$ large enough so that $k>1 / a_{1}$, so $k a_{1}>1$.