I assume that there is no orbital angular momentum, and so we are just keeping track of spin.
First of all, if the particles were distinguishable, we could assign m=-1,0,+1 to each particle (where m is the eigenvalue of S_z for that particle). Then there would be 3^N possible states.
If the particles are bosons, however, then we are allowed to keep only the totally symmetric part of the state. Thus the state is labeled just by the total numbers of +1's, 0's, and -1's, which I will call n_+, n_0, n_-, with n_++n_0+n_-=N. For example, with N=3 and n_+=2, n_0=1, n_-=0, the only allowed state is
{\textstyle{1\over\sqrt3}}(|{+1},+1,0\rangle+|{+1},0,+1\rangle+|0,+1,+1\rangle).
The total number of completely-symmetric states is given by the number of ways to partition N among n_+, n_0, n_-. This number is
\sum_{n_-=0}^N\sum_{n_0=0}^N\sum_{n_+=0}^N \delta_{n_++n_0+n_-,\,N} = \sum_{n_-=0}^N\sum_{n_0=0}^{N-n_-}1 = \sum_{n_-=0}^N N+1-n_- = {\textstyle{1\over2}}(N+1)(N+2).
Suppose we want a state with total S_z=+N. The only way to get it is to have each individual S_z=+1, or equivalently n_+=N, n_0=0, n_-=0. This state must be part of a multiplet with total spin quantum number s=N. There are 2N+1 states in this multiplet, with z-component quantum number ranging from m=-N to m=+N.
Now suppose we want a state with total S_z=N-1. The only way to get it is to have n_+=N-1, n_0=1, n_-=0. However, we already have one state with m=N-1, the one with s=N. So the one state with m=N-1 must be the one in the s=N multiplet. Therefore, there is no state with s=N-1. If there was, there would have to be a second, different state with m=N-1.
Now suppose we want a state with total S_z=N-2. There are two ways to get it: n_+=N-2, n_0=2, n_-=0 and n_+=N-1, n_0=0, n_-=1. One of these states must be the one with s=N and m=N-2, and the other must be a state with s=N-2 and m=N-2. There are 2(N-2)+1=2N-1 states in this multiplet.
Now suppose we want a state with total S_z=N-3. There are two ways to get it: n_+=N-3, n_0=3, n_-=0 and n_+=N-2, n_0=1 , n_-=1. However, we already have two states with m=N-3, the one with s=N and the one with s=N-2. So there is no room for another state with m=N-3, and so there is no state with s=N-3.
And so on. We ultimately find s=N,N-2,N-4,\ldots, down to 1 or 0, are the allowed values.
We can check this by counting states. We already counted the total number of completely symmetric states, with the result {\textstyle{1\over2}}(N+1)(N+2). Now let's count them according to the allowed values of s.
If N is even, the allowed values of s are 0,2,4,\ldots,N, and the total number of states is
\sum_{s=0,2,4,\ldots}^N2s+1 = {\textstyle{1\over2}}(N+1)(N+2),
which agrees with the other way of counting. If N is odd, the allowed values of s are 1,3,5,\ldots,N, and the total number of states is
\sum_{s=1,3,5,\ldots}^N2s+1 = {\textstyle{1\over2}}(N+1)(N+2),
which again agrees.
So we're done. If N is even, the allowed values of s are 0,2,4,\ldots,N, and if N is odd, the allowed values of s are 1,3,5,\ldots,N. In both cases, each value of s occurs once.
