The explanations given by others are horrible. I'll assume you have no knowledge about probability at all.
There is something symmetric going on. What I mean by this is, for example when the test has 3 questions in total, the probability of getting 3 questions correct is equal to getting 0 correct, and the probability of 2 correct is equal to 1 correct. Why? Because getting all 3 correct is just another way of saying getting all 3 wrong. If there is a strange student that aims at answering the questions wrong, the chance that he gets the perfect score (getting all 3 wrong) should be the same as the chance of a normal student aiming to get all 3 right. The same goes for the 2 correct 1 wrong vs 1 correct 2 wrong case.
Now you understand the symmetry, you need to count the total possible outcomes. For example when there are 3 questions, you can get 0 correct, 1 correct, 2 correct, 3 correct. There are 4 total possible outcomes. But when there are 4, you can get 0, 1, 2, 3, or 4 correct. There are 5 total possible outcomes.
Now we are ready to answer your question. How to get the probability of passing when there are 3 questions? Add up the probability of 2 correct and 3 correct. But we know by symmetry, this sum is the same as probability of (0 correct plus 1 correct). The chance of passing is 50%. But when there are 4 questions, the chance of passing is the sum of probability of (2, 3, 4 correct). The chance of failing is the sum of probability of (0, 1 correct). This is obviously less than 50%.
So to conclude, the reason is because when there are even number of questions, the chance of passing is always the chance of failing plus answering a half of the questions correct. While when odd, the chance of both failing and passing is the same.