Fourier serie coefficients of real function
Let me rewrite that statement for you, I believe I know where your misunderstanding lies:


  \$F(n) \space \forall \space (n \in \mathbb{Z}) \$ is purely imaginary and odd if and only if f(t) is odd.


In plain English: \$F(n)\$ (for all n in integers including 0) is purely imaginary and odd [...]. In your example, \$F(0)\$ is not imaginary. Therefore the statement remains correct.

Now think of a new waveform that looks exactly the same as your original "sawtooth" function, but that is now odd: \$g(t) = f(t) - F(0) = f(t) - 1/2\$. If you redo the exercise, you will find \$G(0) = 0\$, confirming the statement.