3. First, prove the necessity:

Let $x_{1}=x_{2}=\cdots=x_{n}=1$, we get $\sum_{i=1}^{n} a_{i} \leqslant \sum_{i=1}^{n} b_{i}$. Let $x_{1}=x_{2}=\cdots=x_{n}=-1$, we get $-\sum_{i=1}^{n} a_{i} \leqslant-\sum_{i=1}^{n} b_{i}$. Therefore, $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$.
Let $x_{1}=x_{2}=\cdots=x_{k}=0, x_{k+1}=x_{k+2}=\cdots=x_{n}=1$ we get $\sum_{i=k+1}^{n} a_{i} \leqslant \sum_{i=k+1}^{n} b_{i}$.
By $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}$, we get $\sum_{i=1}^{k} a_{i} \geqslant \sum_{i=1}^{k} b_{i},(i=1,2, \cdots, n-1)$.
Next, prove the sufficiency: Let $S_{k}=\sum_{i=1}^{k}\left(a_{i}-b_{i}\right),\left(i=1,2, \cdots, n, S_{0}=0\right)$, then $a_{k}-b_{k}=S_{k}-$ $S_{k-1} \cdot(k=1,2, \cdots, n)$ and $S_{k} \geqslant 0 .\left(k=1,2, \cdots, n-1, S_{n}=0\right)$. For any real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1} \leqslant$ $x_{2} \leqslant \cdots \leqslant x_{n}$, we have
$$\begin{aligned} \sum_{i=1}^{n} a_{i} x_{i}-\sum_{i=1}^{n} b_{i} x_{i}= & \sum_{i=1}^{n}\left(a_{i}-b_{i}\right) x_{i}=\sum_{i=1}^{n}\left(S_{i}-S_{i-1}\right) x_{i}= \\ & \sum_{i=1}^{n} S_{i} x_{i}-\sum_{i=1}^{n} S_{i-1} x_{i}= \\ & \sum_{i=1}^{n-1} S_{i} x_{i}-\sum_{i=1}^{n-1} S_{i} x_{i+1}= \\ & \sum_{i=1}^{n-1} S_{i}\left(x_{i}-x_{i+1}\right) \leqslant 0 \end{aligned}$$

That is
$$\sum_{i=1}^{n} a_{i} x_{i} \leqslant \sum_{i=1}^{n} b_{i} x_{i}$$