= Solution
\b[True.] Use the normalization in which the blowup weight $\lambda$ is the area of an exceptional projective line, equivalently the capacity $\pi r^2$ of the ball removed. The auxiliary volume formula for a point <symplectic blowup> in real dimension $2n$ is
$$
\int_{\widetilde X}\frac{\widetilde\omega^n}{n!}
=\int_X\frac{\omega^n}{n!}-\frac{\lambda^n}{n!}.
$$
It applies to each successive blowup, irrespective of where the point is chosen. Geometrically the removed standard ball has volume $\pi^nr^{2n}/n!$, while its replacement divisor has zero $2n$-dimensional volume. If $V$ is the original <symplectic volume>, after $k$ equal-weight blowups the volume is $V-k\lambda^n/n!$. A nonempty closed <symplectic manifold> has strictly positive volume, so
$$
\boxed{k<\frac{n!V}{\lambda^n}.}
$$
This finite upper bound suffices; it does not assert that every number below the bound is realizable. A convention taking the weight to be the radius instead changes the fixed decrement to $\pi^n\lambda^{2n}/n!$, with the same conclusion.
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