= Solution
Choose a subsequence $a_{n_j}$ so sparse that
$$
\sum_{\ell>j}a_{n_\ell}<a_{n_j}
$$
for every $j$; this is possible because the tails of the convergent positive <series> tend to zero. Distinct binary <sequences> $(\varepsilon_j)$ then have distinct sums $\sum_j\varepsilon_ja_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set $\theta_{n_j}=\varepsilon_j$ and all other $\theta_n=0$.
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