Once you grasp that, the answer is simple, you treat the competing dogs as two discrete classes, small and large. One, small dogs, consists of 36 members, and the other, large dogs, consists of 13 members. You now no longer are required to compare the relative size of the small dog class to the large dog class but just determine the fact of how many more there are above the total there may be, regardless of the large dog class’ number. The reading comprehension enters in ignoring the extraneous data which is mere unnecessary "noise" to respond to the question. The problem answers itself by providing the answer, 36 small dogs.
This avoids the problem of chopped wiener dog. . .
Now, if you really want to make it a maths problem, then one has to realize the issue is significant digits. . . that one is working with non-divisible units, incapable of being divided further into fractions. I.e. absolute numbers: for example, one cannot subtract 5 dogs from 3 dogs and have a minus 2 dogs. It’s not possible in the real world. . . Just as in this case one cannot enter half-dog in a dog show.
The teacher who originally wrote this problem was stupid. He or she should have worked the math first. I do note this was a second grade problem for a seven year old, who is not yet ready to start algebra. I think the problem was a typo. The question for a word problem at this grade level would have been: “How many large dogs are competing?"
This is a good example of why math does NOT represent reality, say like climate change models, can’t really predict the future weather.
You are exactly right Swordmaker, and provide an excellent analysis of the problem.