Three. (50 points) Find the smallest positive integer $n$, such that there exist rational numbers $a_{i}, b_{i} (i=1,2, \cdots, n)$ satisfying
$$
x^{2}+x+4=\sum_{i=1}^{n}\left(a_{i} x+b_{i}\right)^{2} .
$$
Three, first, from
$$
\begin{array}{l}
x^{2}+x+4 \\
=\left(x+\frac{1}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}+1^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2},
\end{array}
$$

we know the smallest positive integer $n \leqslant 5$.
Next, prove: for all $n \geqslant 5$.
Assume there exist rational numbers $a_{i} 、 b_{i}(i=1,2,3,4)$, such that
$$
x^{2}+x+4=\sum_{i=1}^{4}\left(a_{i} x+b_{i}\right)^{2} \text {. }
$$

By comparing coefficients, we get
$$
\left\{\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}=1, \\
b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}=4, \\
a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+a_{4} b_{4}=\frac{1}{2} .
\end{array}\right.
$$

Then $\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\right)=4$.
By the product of four squares is still a sum of four squares, i.e.,
$$
\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\right) \\
=\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+a_{4} b_{4}\right)^{2}+ \\
\left(a_{1} b_{2}-a_{2} b_{1}+a_{3} b_{4}-a_{4} b_{3}\right)+ \\
\left(a_{1} b_{3}-a_{3} b_{1}+a_{4} b_{2}-a_{2} b_{4}\right)^{2}+ \\
\left(a_{1} b_{4}-a_{4} b_{1}+a_{2} b_{3}-a_{3} b_{2}\right)^{2},
\end{array}
$$

Thus, there exist rational numbers $x_{i} \in \mathbf{Q}(i=1,2,3)$, such that
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=4-\frac{1}{4}=\frac{15}{4} \\
\Rightarrow\left(2 x_{1}\right)^{2}+\left(2 x_{2}\right)^{2}+\left(2 x_{3}\right)^{2}=15 .
\end{array}
$$

Clearing the denominators and rearranging, we get there exist $a 、 b 、 c 、 d \in \mathbf{Z}$, and $(a, b, c, d)=1$, such that
$$
a^{2}+b^{2}+c^{2}=15 d^{2} ;
$$

Taking modulo 16, we get
$$
a^{2}+b^{2}+c^{2}+d^{2} \equiv 0(\bmod 8) \text {. }
$$

It is easy to see that the number of even numbers among $a 、 b 、 c 、 d$ can be 0 or 2. If there are no even numbers among $a 、 b 、 c 、 d$, then
$$
4 \equiv a^{2}+b^{2}+c^{2}+d^{2} \equiv 0(\bmod 8),
$$

which is a contradiction;
If there are exactly two even numbers among $a 、 b 、 c 、 d$, then
$$
4\left(u^{2}+v^{2}\right)+2 \equiv a^{2}+b^{2}+c^{2}+d^{2} \equiv 0(\bmod 8) \text {, }
$$

which is a contradiction.
Therefore, $x^{2}+x+4$ cannot be expressed as the sum of squares of four rational linear expressions.
Thus, $n \geqslant 5$.
In summary, the smallest positive integer $n=5$.