= Solution
Starting with $\theta_0=\theta$, define
$$
a_n=\lfloor\theta_n\rfloor,
\qquad
\theta_{n+1}=\frac1{\theta_n-a_n}
$$
while the denominator is nonzero. This gives the <continued fraction>[simple continued fraction]
$$
\theta=[a_0;a_1,a_2,\ldots].
$$
A finite expansion is rational by evaluating it from the bottom. Conversely, for rational $\theta$, these steps are the <Euclidean algorithm> applied to numerator and denominator, so the remainders eventually vanish.
Define convergents by
$$
p_{-2}=0, p_{-1}=1,quad q_{-2}=1, q_{-1}=0,
$$
$$
p_n=a_np_{n-1}+p_{n-2},\qquad
q_n=a_nq_{n-1}+q_{n-2}.
$$
The recurrence gives
$$
p_nq_{n-1}-p_{n-1}q_n
=-(p_{n-1}q_{n-2}-p_{n-2}q_{n-1}),
$$
so induction yields
$$
\boxed{p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}}.
$$
In particular,
$$
\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right|
=\frac1{q_nq_{n+1}}.
$$
Since an irrational $\theta$ lies strictly between these convergents, the two approximation errors sum to this distance. If both displayed bounds in the question failed, their sum would be at least
$$
\frac1{2q_n^2}+\frac1{2q_{n+1}^2}
\geq\frac1{q_nq_{n+1}},
$$
by the arithmetic-geometric mean inequality, contradicting strict betweenness. Thus at least one bound holds.
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