Set $x_i = \theta_i^2+b_i$, $1\le i\le n$. We obtain
\[
\sum_{i\le n}c_i \theta_i = \frac12 \sum_{i\le n}\bigl(\theta_i^2 + b_i\bigr) \iff \sum_{i\le n} \bigl(\theta_i-c_i\bigr)^2 = \sum_i \bigl(c_i^2-b_i\bigr)\triangleq S.
\]Note that if $S=0$, then we indeed have a unique solution: $x_i = c_i^2+b_i$. If $S<0$, we obtain no solutions, whereas for $S>0$, it is not hard to see the solutions are not unique: (a) $x_1=b_1+(c_1+\sqrt{S})^2$, $x_i=b_i+c_i^2$, $i\ge 2$ and (b) $x_i = b_i+(c_i+\sqrt{S/2})^2, i\in\{1,2\}$ and (b) $x_i = b_i+c_i^2$ for $3\le i\le n$ are distinct solutions.