Upper bound: A quadrature rule with $n$ nodes has $2n$ free parameters (nodes and weights). Since a polynomial of degree $2n$ has $2n + 1$ coefficients, one cannot expect exactness beyond degree $2n - 1$ in general.

Existence and characterization: Let $p_0, p_1, \ldots$ be the orthonormal polynomials with respect to $\langle f, g \rangle = \int_a^b w(x) f(x) g(x)\, dx$. Suppose the nodes $x_1, \ldots, x_n$ are the roots of $p_n(x)$.

For any polynomial $f$ of degree $\leq 2n - 1$, divide by $p_n$: $f(x) = q(x) p_n(x) + r(x)$ where $\deg q \leq n - 1$ and $\deg r \leq n - 1$.

Since $\deg q \leq n - 1$ and $p_n \perp p_k$ for $k < n$: $\int_a^b w(x) q(x) p_n(x)\, dx = 0$.

Therefore $\int_a^b w(x) f(x)\, dx = \int_a^b w(x) r(x)\, dx$.

Since $p_n(x_i) = 0$ for all $i$, we have $f(x_i) = r(x_i)$, so $\sum_{i=1}^n w_i f(x_i) = \sum_{i=1}^n w_i r(x_i)$.

The weights are chosen so that the quadrature is exact for all polynomials of degree $\leq n - 1$ (which is possible since the Lagrange interpolation weights provide the unique such choice). Since $\deg r \leq n - 1$: $\sum_{i=1}^n w_i r(x_i) = \int_a^b w(x) r(x)\, dx$.

Hence $\sum_{i=1}^n w_i f(x_i) = \int_a^b w(x) f(x)\, dx$ for all $f$ with $\deg f \leq 2n - 1$.

Uniqueness of nodes: If $x_1, \ldots, x_n$ achieve degree of exactness $2n - 1$, define $\omega(x) = \prod_{i=1}^n (x - x_i)$. For any polynomial $q$ with $\deg q \leq n - 1$: $\int_a^b w(x) q(x)\omega(x)\, dx = \sum_{i=1}^n w_i q(x_i)\omega(x_i) = 0$ since $\omega(x_i) = 0$. Thus $\omega \perp$ all polynomials of lower degree, so $\omega$ is (up to scaling) the orthogonal polynomial $p_n$. $\square$