Lots of cats and rats

From hole #1 = n
Eaten by m cats (one each), - m
Into hole #1 = n - m

From hole #2 = 2(n - m)
Eaten by m cats (one each), - m
Into hole #2 = 2n - 3m

From hole #3 = 2(2n - 3m)
Eaten by m cats (one each), - m
Into hole #3 = 4n - 7m

From hole #4 = 2(4n - 7m)
Eaten by m cats (one each), - m
Into hole #4 = 8n - 15m = 0
8n = 15m
m = 8k and n = 15k
Taking k = 1 for the smallest solution,

Answer = m + n = 8 + 15 = 23

We know that for there to have been no survivors, $m$ rats came out of the $4^{th}$ hole, which means $\frac{m}{2}$ went into the third hole. Repeating this process, $\frac{3m}{2}$ rats came out of the $3^{rd}$ hole, and $\frac{3m}{4}$ rats came went back into the second hole. Then, we know that $\frac{7m}{4}$ rats came out of the second hole, and $\frac{7m}{8}$ went into the first. Finally, we know that there were originally $\frac{15m}{8}$ rats in the first hole. Because we know m and n are coprime, m must equal 8 and n equals 15. Lastly, $8+15=\boxed{23}$

15 rats and 8 cats.

1st hole- 15 rats, surviving rats =7

2nd hole-14 rats, surviving rats =6

3rd hole-12 rats,surviving rats =4

4th hole-8 rats, surviving rats=0