SOLUTION: The sum of the digits of a two-digit number is 9. One-half of the number is also a two-digit number whose sum of digits is also 9. How many such numbers exist?

Let t = the tens digit of one-half the number.
Let u = the units digit of on-half the number.

10t+u = one-half the number
2(10t+u) = the number

t+u = 9
  u = 9-t

2[10t+(9-t)] = the number

2[10t+9-t] = the number

2[9t+9] = the number

2 = the number

18(t+1) = the number

10 <= 18(t+1) <= 99 

.5... <= t+1 <= 5.5

Subtract 1 from all 3 sides

.4... < t < 4.5

The tens digit of one-half the number is 1,2,3 or 4

Since the sum of the digits of one-half the number is 9,
the tens digits are 8,7,6, or 5

So one half the number is 18, 27, 36, or 45

So the number is either 36, 54, 72, or 90. 

Edwin