3. Let the polynomial be
$$
f(x)=x^{2020}+\sum_{i=0}^{2019} c_{i} x^{i}\left(c_{i} \in\{-1,0,1\}\right) \text {. }
$$

Let $N$ be the number of positive integer roots of $f(x)$ (including multiplicities). If the equation $f(x)$ has no negative integer roots, find the maximum value of $N$.
（Luo Hua
3. First, since $c_{i} \in\{-1,0,1\}$, the absolute value of the roots of $f(x)$ cannot be greater than or equal to 2. Therefore, only the integer roots $-1$, $0$, and $1$ need to be considered.
Notice,
$$
\begin{aligned}
f(x)= & (x-1)\left(x^{3}-1\right)\left(x^{5}-1\right)\left(x^{11}-1\right) . \\
& \left(x^{21}-1\right)\left(x^{43}-1\right)\left(x^{85}-1\right)\left(x^{171}-1\right) . \\
& \left(x^{341}-1\right)\left(x^{683}-1\right) x^{656},
\end{aligned}
$$

satisfies the condition.
At this point, the degree of $f(x)$ is 2020, where the number of roots $x=1$ is $N=10$, and $-1$ is not a root of $f(x)$.
Let $f_{i}=x^{q_{i}}-1$, where,
$q_{1}=1, q_{2}=3, q_{3}=5, q_{4}=11, q_{5}=21, q_{6}=43$,
$q_{7}=85, q_{8}=171, q_{9}=341, q_{10}=683$.
It is easy to prove by induction that $\prod_{i=1}^{J} f_{i} (J=1,2, \cdots, 10)$ is

a polynomial with all coefficients in $\{-1,0,1\}$.
Next, prove that $N=10$ is the maximum value.
Assume $N \geqslant 11$. Then
$$
\begin{array}{l}
(x-1)^{11} \mid\left(x^{2020}+\sum_{i=0}^{2019} c_{i} x^{i}\right) . \\
\text { Let } x=-1 \text {, then } 2^{11} \mid f(-1) .
\end{array}
$$

Since $|f(-1)| \neq 0$, we have $|f(-1)| \geqslant 2^{11}$. At the same time,
$$
|f(-1)| \leqslant 1+\sum_{i=0}^{2019} c_{i} \leqslant 2021<2^{11} \text {, }
$$

which is a contradiction. Therefore, the maximum value of $N$ is 10.