Let $V$ be the volume of the first tank.

$\dfrac{V}{2} ; \frac{V}{4} ; \dfrac{V}{8} \ldots$

$S_{19}=\dfrac{\dfrac{V}{2}\left[1-\left(\dfrac{1}{2}\right)^{19}\right]}{1-\dfrac{1}{2}}$

$=\dfrac{524287}{524288} V$

$=0,9999980927 V$

$<V$

Yes, the water will fill the tank without spilling

**OR**

Let $V$ be the volume of the first tank.

$\dfrac{V}{2} ; \dfrac{V}{4} ; \dfrac{V}{8} \ldots$

$S_{19}=\dfrac{\dfrac{V}{2}\left[1-\left(\dfrac{1}{2}\right)^{19}\right]}{1-\dfrac{1}{2}}$

$=V\left[1-\left(\frac{1}{2}\right)^{19}\right]$

$<V .1$

$<V$

Yes, the water will fill the tank without spilling

**OR**

Let $V$ be the volume of the first tank.

$\dfrac{V}{2} ; \dfrac{V}{4} ; \dfrac{V}{8} \ldots$

$S_{\infty}=\dfrac{\dfrac{V}{2}}{1-\dfrac{1}{2}}$

$=V$

Since the first tank will hold the water from infinitely many tanks without spilling over, certainly:

Yes, the first tank will hold the water from the other 19 tanks without spilling water.

**OR**

If the tanks are emptied one by one, starting from the second, each tank will fill only half the remaining space, so the first tank can hold all the water from the other 19 tanks.