3. Find the smallest positive integer $n$, such that there exist rational coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
3. Since $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$, then $n \leqslant 5$.
Thus, it only needs to be proven that $x^{2}+7$ is not equal to the sum of squares of no more than four rational coefficient polynomials.

Assume there exist four rational coefficient polynomials $f_{1}, f_{2}, f_{3}, f_{4}$ (some of which may be 0), satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+f_{3}^{2}(x)+f_{4}^{2}(x).
$$

Then the degrees of $f_{1}, f_{2}, f_{3}, f_{4}$ are at most 1.
Let $f_{i}(x)=a_{i} x+b_{i}\left(a_{i}, b_{i} \in \mathbf{Q}, i=1,2,3,4\right)$.
From $x^{2}+7=\sum_{i=1}^{4}\left(a_{i} x+b_{i}\right)^{2}$, we get
$$
\sum_{i=1}^{4} a_{i}^{2}=1, \sum_{i=1}^{4} a_{i} b_{i}=0, \sum_{i=1}^{4} b_{i}^{2}=7.
$$

Let $p_{i}=a_{i}+b_{i}, q_{i}=a_{i}-b_{i}(i=1,2,3,4)$.
Then $\sum_{i=1}^{4} p_{i}^{2}=\sum_{i=1}^{4} a_{i}^{2}+2 \sum_{i=1}^{4} a_{i} b_{i}+\sum_{i=1}^{4} b_{i}^{2}=8$,
$\sum_{i=1}^{4} q_{i}^{2}=\sum_{i=1}^{4} a_{i}^{2}-2 \sum_{i=1}^{4} a_{i} b_{i}+\sum_{i=1}^{4} b_{i}^{2}=8$,
$\sum_{i=1}^{4} p_{i} q_{i}=\sum_{i=1}^{4} a_{i}^{2}-\sum_{i=1}^{4} b_{i}^{2}=-6$.
This implies there exist integers $x_{i}, y_{i}(i=1,2,3,4)$, $m(m>0)$ satisfying the following equations
(1) $\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}$,
(2) $\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}$,
(3) $\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2}$.
Assume the above equations have a solution, consider a solution that makes $m$ the smallest.
Note that, when $x$ is odd, $x^{2} \equiv 1(\bmod 8)$; when $x$ is even,
$x^{2} \equiv 0(\bmod 8)$ or $x^{2} \equiv 4(\bmod 8)$.
From (1), we know that $x_{1}, x_{2}, x_{3}, x_{4}$ are all even, and from (2), we know that $y_{1}, y_{2}, y_{3}, y_{4}$ are also all even. Thus, the left side of (3) can be divided by 4. Therefore, $m$ is even.

Hence $\left(\frac{x_{1}}{2}, \frac{y_{1}}{2}, \frac{x_{2}}{2}, \frac{y_{2}}{2}, \frac{x_{3}}{2}, \frac{y_{3}}{2}, \frac{x_{4}}{2}, \frac{y_{4}}{2}, \frac{m}{2}\right)$ is also a solution to the above equations, which contradicts the minimal selection of $m$.
In conclusion, the minimum value of $n$ is 5.