To find the maximum number of positive integer roots $N$ of the polynomial $f(x) = x^{2020} + \sum_{i=0}^{2019} c_i x^i$ where $c_i \in \{-1, 0, 1\}$, and given that $f(x) = 0$ has no negative integer roots, we proceed as follows:

1. Construct a Polynomial with Positive Integer Roots:
Consider the polynomial $f(x) = (x-1)(x^3-1)(x^7-1)(x^{13}-1)(x^{25}-1)(x^{49}-1)(x^{99}-1)(x^{199}-1)(x^{399}-1)(x^{799}-1)x^{426}$.

- This polynomial has roots at $x = 1$ with multiplicity 1, $x = \sqrt[3]{1}$ with multiplicity 3, $x = \sqrt[7]{1}$ with multiplicity 7, and so on.
- The exponents $1, 3, 7, 13, 25, 49, 99, 199, 399, 799$ are chosen such that their sum plus the exponent 426 equals 2020.

2. Verify the Polynomial:
- The polynomial $f(x)$ constructed above is valid because it has the form $x^{2020} + \sum_{i=0}^{2019} c_i x^i$ with $c_i \in \{-1, 0, 1\}$.
- The polynomial has 10 positive integer roots: $1, \sqrt[3]{1}, \sqrt[7]{1}, \sqrt[13]{1}, \sqrt[25]{1}, \sqrt[49]{1}, \sqrt[99]{1}, \sqrt[199]{1}, \sqrt[399]{1}, \sqrt[799]{1}$.

3. Prove the Maximum Number of Positive Integer Roots:
- Suppose $N > 10$. Then $N \geq 11$.
- If $N \geq 11$, then $(x-1)^{11}$ must divide $f(x)$.
- Evaluate $f(-1)$:
$$f(-1) = (-1)^{2020} + \sum_{i=0}^{2019} c_i (-1)^i = 1 + \sum_{i=0}^{2019} c_i (-1)^i$$
- Since $c_i \in \{-1, 0, 1\}$, the maximum absolute value of $f(-1)$ is:
$$|f(-1)| \leq 1 + \sum_{i=0}^{2019} 1 = 2021$$
- However, if $(x-1)^{11}$ divides $f(x)$, then $2^{11}$ must divide $f(-1)$, implying:
$$|f(-1)| \geq 2^{11} = 2048$$
- This is a contradiction because $2021 < 2048$.

Therefore, the maximum number of positive integer roots $N$ is 10.

The final answer is $\boxed{ 10 }$.