It is not mentioned, so I assume that prisoners can escape from the prison from any cell. Also, they can only escape in groups of 4 at a time.
First, all the prisoners will divide themselves up into 9/cell. Amongst themselves, they will all order themselves. Now, we will pick two opposite corner cells - lets say the top left and bottom right. Unfortunately for them, these prisoners are stuck and will never get free. We will label them as 'P' for 'Permanent prisoner'. The other corners we will label a-i. The sides are numbered 1-9. So we have the following:
PPP|123|abc
PPP|456|def
PPP|789|ghi
---+---+---
123| |123
456| |456
789| |789
---+---+---
abc|123|PPP
def|456|PPP
ghi|789|PPP
More concisely:
Px9 | 1-9 | a-i
-----+-----+-----
1-9 | | 1-9
-----+-----+-----
a-i | 1-9 | Px9
Now that the prisoners have numbered themselves, the following things will happen simultaneously:
- The 2 lowest lettered prisoners in the corners (a&b) escape!
- The 2 lowest numbered prisoners in the sides (1&2) shift towards the nearest corner.
Then we have:
PPP1| 3| c
PPP2|456|def
PPP |789|ghi
12 | |
----+---+---
3 | | 3
456 | |456
789 | |789
----+---+---
c | 3|PPP1
def |456|PPP2
ghi |789|PPP
| |12
Or more concisely:
Px13 | 3-9 | c-i
-----+-----+-----
3-9 | | 3-9
-----+-----+-----
c-i | 3-9 | Px13
Each row and column still contains 27 prisoners; $13+7+7 = 9+9+9 = 27$.
Repeat this process 3 more times. Then we will have;
Px25 | 9 | i
-----+-----+-----
9 | | 9
-----+-----+-----
i | 9 | Px25
There are still 27 prisoners in each row and column ($25+1+1=27$), and $4
\times 4 = 16$ prisoners have escaped.