[Proof] Proposition Proof:
$$\begin{aligned} & a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \\ = & \left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \\ = & \sum_{i, j=1}^{n} a_{i} x_{i} x_{j} \\ = & \sum_{i=1}^{n} a_{i} x_{i}^{2}+\sum_{1 \leqslant i<j \leqslant n}\left(a_{i}+a_{j}\right) x_{i} x_{j} \\ \geqslant & a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2} . \end{aligned}$$

Statement of the Converse: Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers. If for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying
$$x_{1}+x_{2}+\cdots+x_{n}=1$$

we have
$$a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \geqslant a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2}$$

then the sum of any two numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is non-negative.
Proof of the Converse: Let $i, j$ be natural numbers such that
$$1 \leqslant i \leqslant n, 1 \leqslant j \leqslant n, i \neq j$$

Take $x_{1}, x_{2}, \cdots, x_{n}$ as: $x_{i}=x_{j}=\frac{1}{2}, x_{k}=0(k \neq i, k \neq j)$.
From $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \geqslant a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2}$, we get $\quad \frac{1}{2}\left(a_{i}+a_{j}\right) \geqslant \frac{1}{4}\left(a_{i}+a_{j}\right)$,
which implies $a_{i}+a_{j} \geqslant 0$.