I know it has something to do with 2's complement and adding 1, but I don't really get how you can encode one more number with the same amount of bits when it comes to negative numbers.
Think about it in these terms. Take a 2-bit number with a preceding sign:
Now let's have some negatives:
Wait, we also have
It has to be negative, because the sign-bit is 1. So, logically, it must be -4.
(Edit: As WorldEngineer rightly points out, not all numbering systems work this way -- but the ones you're asking about do.)
There are BASICALLY three ways to represent signed integers in binary: 2's complement, 1's complement, and sign-magnitude. (Biquinary went the way of the Dodo Bird a long time ago.)
1's complement and sign-magnitude have two zero values, +0 and -0, each with a unique representation. 2's complement only has one zero value, and one representation.
Now, a field of N bits can encode 2^N values. Subtract one in 2's complement, and you have 2^N-1 = 2^(N-1) + 2^(N-1) + 1. Since the representation for zero is all zero bits, and a + sign is zero, there will be one more possible nonzero representation with the sign bit set to 1.
This is a very long-winded way of saying 2's complement represents values in the range -(2^(N-1)) .. +(2^(N-1) - 1).
1's complement actually has an advantage over 2's complement if you are doing integer digital signal processing computations. 1's complement operations inherently truncate toward zero. 2's complement truncates toward -infinity. I learned this one the HARD way...
Because there are not two classes of numbers in the integer range, but three: negative numbers, zero, and positive numbers. Zero has to take up a slot (would be rather impractical not to be able to represent zero...), so either the positive or the negative class has to give up a slot. The fact that it's usually the positive range that has to make that sacrifice is to a certain extent arbitrary, but on the level of bit manipulations there are some things that this decision makes more convenient.