= Solution
The <Legendre polynomial> $p_n$ is <orthogonal polynomial>[orthogonal] on $[-1,1]$ to every polynomial of degree below $n$. Suppose it changes sign at only $r<n$ distinct interior points $x_1,\ldots,x_r$, and put
$$
q(x)=\prod_{j=1}^r(x-x_j),
$$
choosing its overall sign so that $p_nq\ge0$ on $[-1,1]$. This product is positive except at finitely many points, so
$$
\int_{-1}^1p_n(x)q(x)\,dx>0,
$$
contradicting orthogonality because $\deg q=r<n$. Thus $p_n$ has at least $n$ sign-changing roots in $(-1,1)$; its degree is $n$, so these are all its roots and they are distinct.
For arbitrary distinct nodes $x_1,\ldots,x_n$, let
$$
L_i(t)=\prod_{j\ne i}\frac{t-x_j}{x_i-x_j}
$$
be the <Lagrange interpolation polynomial>[Lagrange basis polynomials]. Every polynomial $P$ of degree below $n$ satisfies $P(t)=\sum_iP(x_i)L_i(t)$, so the required weights exist and are
$$
A_i=\int_{-1}^1L_i(t)\,dt.
$$
They are unique because applying any such formula to $P=L_i$ recovers $A_i$.
Now assume exactness for every $P$ of degree below $2n$ and define $Q(t)=\prod_i(t-x_i)$. For every polynomial $R$ of degree below $n$, the product $QR$ has degree below $2n$ and vanishes at every node. Therefore
$$
\int_{-1}^1Q(t)R(t)\,dt=0.
$$
There is a unique monic polynomial of degree $n$ orthogonal to all lower-degree polynomials, so $Q$ is the monic normalization of $p_n$ and the $x_i$ are precisely the roots of $p_n$. Taking $P=1$ gives
$$
\boxed{\sum_{i=1}^nA_i=2}.
$$
Finally $L_i^2$ has degree $2n-2$, so exactness gives
$$
A_i=\sum_jA_jL_i(x_j)^2
=\int_{-1}^1L_i(t)^2\,dt>0.
$$
These are the positive weights of <Gaussian quadrature>.
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