Solution: As shown on the right, if the cross-sectional DEM is parallel to the bottom ABC, then m is the midpoint of SC and f is the midpoint of SM. If the cross section FNP is parallel to the bottom ABC, then n and p are the midpoint of SD and SE, respectively.

Let the volume of triangular pyramid S-ABC be V and the height h, and the volume of S-DEM be V 1 and the height h, then hH=23, V 1V = (23) 3 = 827.

The ratio of the volume of triangular pyramid F-DEM to that of triangular pyramid S-DEM is 1: 2 (high ratio),

∴ volume of triangular pyramid F-DEM 4V27 ... volume of triangular prism DEM-ABC = v-v1=19v27,

∴ Maximum water capacity = 4V27+ 19V27 = 23V27.

So the maximum volume of water is 2327.

So the answer is: 2327.