 Solution

A defect of charge $q$ has far-field elastic energy proportional to $q^2\log (L/r_0)$. Splitting $q$ into allowed charges $q_i$ with $\sum q_i=q$ lowers $\sum q_i^2$ unless the pieces already have the smallest permitted magnitude. Polar order distinguishes $\mathbf{p}$ from $-\mathbf{p}$, so its smallest nonzero charge is $|q|=1$. A nematic liquid crystal identifies $\hat{\mathbf{n}}\sim -\hat{\mathbf{n}}$; a rotation by only $\pi$ closes a loop in its order-parameter space, permitting $q=\pm 1/2$.