That one cannot guarantee a zero on $(0,\pi/\delta]$ or on any
interval $(0,c]$ with $c$ independent on $n$ follows from the example
of B. Logan, (Theorem 5.5.1 of his thesis Properties of high-pass signals, Columbia Univ., 1965). He constructs a bounded $L^2$ function whose Fourier transform is supported on any given set $[-b,-a]\cup [a,b]$ with $0<a<b$,
and which is positive on any given interval $(0,c)$. This function can be approximated by functions of your class with large $n$.

Logan's example is reproduced as Example 1 here.

Here is the solution suggested by user Fedja.

Theorem. $f$ must have a zero on the interval $(0,\pi n/\delta)$.
(This estimate is unlikely to be exact for all $n$).

Proof. Define the linear operators $f\mapsto A_kf$, where
$(A_kf)(t)=f(t)+f(t+\pi/\omega_k)$. When we apply $A_k$ to our $f$ it kills
all summands of the form $\sin\omega_k(t+\alpha)$. So
$$A_nA_{n-1}\ldots A_1f=0.$$
On the other hand $$(A_nA_{n-1}\ldots A_1f)(t)=f(t)+f(t+\pi/\omega_1)+f(t+\pi/\omega_2)+f(t+\pi/\omega_1+\pi/\omega_2)+\ldots,$$
all summands are of the form $f(t+\alpha)$ where $\alpha$ is a sum of some $\pi/\omega_j$, with at most $n$ such summands,
so $\alpha\leq \pi n/\delta$.
Thus if our $f(t)>0$ on $(0,\pi n/\delta+\epsilon)$ this sum will be positive
on $(0,
\epsilon)$ - contradiction.

Notice that the estimate does not depend on $\Delta$.

EDIT. For $f(x)=(2/3)\sin x+(1/3)\sin(2.5 x)$ the smallest positive zero is about $3.48>\pi$,