According to known conditions, (10c+d)+(10a+10c+d)+(100a+100b+/kloc+d) = 2999.

Finishing:1000a+100 (a+b)+30c+3d = 2992.

(Here, a, b, c and d are all positive integers of one digit by default. In fact, it can also be proved that abcd is a positive integer of one digit. I don't want to prove it here. I want to prove it before asking. )

3d can only be =2, 12 and 22. Obviously, d=4 here.

The above formula becomes1000a+100 (a+b)+30c = 2980.

30c can only be =80,180,280, so it appears that c=6.

The above formula becomes1000a+100 (a+b) = 2800.

Obviously here a=2.

Then a+b=8 and b=8-2=6.

The answer is: I =2, love =6, number =6, learning =4.