I can’t hear the video from this computer, but putting batteries in series increases voltage, but also increases the total resistance.
If each battery has internal resistance of 0,1ohm, 244 of them in series will produce 2196 volts with a 24.4 ohm resistance and that will cause 2196/24.4= 90Amps current when shorted.
If you had half the batteries in series you would get 1098/12,2= 90Amps again!
But if you connected the two 122 series groups in parallel you’d get 1098/6.1= 180Amps
So you will get maximum current using some combination of series/parallel instead of series.
Put them all in parallel. 1/R = 1/r1 + 1/r2 + … 1/rn. In this case all the resistances are 0.1 ohm so for 244 of them in parallel we have 1/R = 244/0.1 and hence R = 1/2440. Then I = V/R gives us a current of 9/(1/2440) = 9 x 2440 = 21,960 A.
If a 9 volt battery could supply 90 amps, it would melt down from a short in seconds, as in the terminals going white-hot (more realistically, 5 amps is about what you’d get; I also tried placing a current meter across a good 9 v battery and the current fell over a few seconds to 2-3 amps, indicating that the DC resistance is even higher that the 1 kHz impedance, see my previous post including a datasheet of the actual internal resistance, around 1.7 ohms).
If you look at the video in Dog80’s post, the beginning shows them shorting out a single 9v battery, with only a tiny spark.
While not exactly the same size (slightly shorter), you can open up a 9v battery and extract the cells and use them in place of AAAA batteries (note that some 9v batteries use flat cells, which appear to be more volumetrically efficient, can’t use them individually though since they are held together and sealed by a plastic wrapper).
