Example 3 Real-coefficient polynomials $f_{i}(x)=a_{i} x^{2}+b_{i} x+c_{i}$ $\left(a_{i}>0, i=1,2, \cdots, 2011\right)$, and $\left\{a_{i}\right\} 、\left\{b_{i}\right\} 、\left\{c_{i}\right\}$ are all arithmetic sequences. If $F(x)=\sum_{i=1}^{204} f_{i}(x)$ has real roots, then at most how many polynomials in $\left\{f_{i}(x)\right\}$ have no real roots?
【Analysis】From $\left\{a_{i}\right\}$ being an arithmetic sequence, we know
$$
\sum_{i=1}^{2011} a_{i}=(2 \times 1005+1) a_{1006}=2011 a_{1006} \text {. }
$$

Similarly, $\sum_{i=1}^{2011} b_{i}=2011 b_{1006}$,
$$
\sum_{i=1}^{2011} c_{i}=2011 c_{1006} \text {. }
$$

Therefore, $F(x)=\sum_{i=1}^{200} f_{i}(x)=2011 f_{1006}(x)$ has a real root, which means $f_{1006}(x)$ has a real root, denoted as $x_{0}$.
When $1 \leqslant k \leqslant 1005$, note that,
$$
f_{k}(x)+f_{2012-k}(x)=2 f_{1006}(x)
$$

has a real root $x_{0}$, i.e.,
$$
f_{k}\left(x_{0}\right)+f_{2012-k}\left(x_{0}\right)=0,
$$

then at least one of $f_{k}\left(x_{0}\right)$ and $f_{2012-k}\left(x_{0}\right)$ is less than or equal to zero.

Thus, at least one of $f_{k}\left(x_{0}\right)$ and $f_{2012-k}\left(x_{0}\right)$ has a real root.

Therefore, at least 1006 of $\left\{f_{i}(x)\right\}$ have real roots, meaning at most 1005 do not have real roots.

Next, we provide a specific example where 1005 of $\left\{f_{i}(x)\right\}$ do not have real roots:
For $f_{i}(x)=x^{2}+i-1006$,
when $1 \leqslant i \leqslant 1006$, $f_{i}(x)$ has real roots; when $1007 \leqslant i \leqslant 2012$, $f_{i}(x)$ does not have real roots.
【Note】This problem uses the extended form of the middle term formula
$$
2 a_{m}=a_{m-k}+a_{m+k} .
$$