The 87th question: Let the polynomial $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2020}+\sum_{\mathrm{i}=0}^{2019} \mathrm{c}_{\mathrm{i}} \mathrm{x}^{\mathrm{i}}\left(\mathrm{c}_{\mathrm{i}} \in\{-1,0,1\}, 0 \leq \mathrm{i} \leq 2019\right)$. Let $\mathrm{N}$ be the number of positive integer roots of $f(x)$ (including multiple roots). If the equation $f(x)$ has no negative integer roots, find the maximum value of $N$.

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The 87th question,
Solution: We first prove a lemma: If the complex number $\mathrm{a}$ satisfies $\mathrm{f}(\mathrm{a})=0$, then $|\mathrm{a}|<2$.
Proof of the lemma: If $|\mathrm{a}| \geq 2$, by $\mathrm{f}(\mathrm{a})=0$ we know:
$$
\begin{array}{l}
\mathrm{a}^{2020}=-\sum_{i=0}^{2019} \mathrm{c}_{\mathrm{i}} \mathrm{a}^{\mathrm{i}} \\
\Rightarrow 1=-\sum_{\mathrm{i}=0}^{2019} \mathrm{c}_{\mathrm{i}} \mathrm{a}^{\mathrm{i}-2020} \\
\Rightarrow 1=\left|-\sum_{\mathrm{i}=0}^{2019} \mathrm{c}_{\mathrm{i}} \mathrm{a}^{\mathrm{i}-2020}\right| \leq \sum_{\mathrm{i}=0}^{2019}\left|\mathrm{c}_{\mathrm{i}}\right| \frac{1}{|\mathrm{a}|^{2020-i}} \leq \sum_{i=0}^{2019} \frac{1}{2^{2020-i}}=1-\frac{1}{2^{2020}}<1
\end{array}
$$

This leads to a contradiction. The lemma is proved!
According to the lemma, and noting that $\mathrm{f}(\mathrm{x})$ has no negative integer roots, the integer roots of $\mathrm{f}(\mathrm{x})$ can only be 0 or 1.
If $N \geq 11$, then $(x-1)^{11} \mid f(x)$, thus $2^{11} \mid f(-1)$. However, $f(-1) \neq 0$, and $|f(-1)| \leq 1+$ $\sum_{i=0}^{2019}\left|c_{i}\right| \leq 2021<2^{11}$, which leads to a contradiction. Therefore, $\mathrm{N} \leq 10$.
On the other hand, take
$$
f(x)=(x-1)\left(x^{3}-1\right)\left(x^{5}-1\right)\left(x^{11}-1\right)\left(x^{21}-1\right)\left(x^{43}-1\right)\left(x^{85}-1\right)\left(x^{171}-1\right)\left(x^{341}-\right.
$$
1) $\left(x^{683}-1\right) x^{656}$

Then $\mathrm{f}(\mathrm{x})$ satisfies the conditions of the problem, and $\mathrm{N}=10$.
In conclusion, the maximum value of $\mathrm{N}$ is 10.