$$ \text{Given } \{n, \thinspace N\} \in \mathbb{N}^+ \text{ and } N > n \\ \ \\ \begin{align} p(x) &= \sum_{n}^N a_n{x^n} \\ \ \\ r(x) &= \sum_{n}^N \frac{a_n}{x^n} \end{align} $$
The highest power $N$ of a polynomial $p(x)$ is called its "degree". What is the highest power $n$ of $r(x)$ called?
First some definitions: A rational function is any function that can be put in the form $f(x)/g(x)$ for polynomials $f(x)$ and $g(x)$. The degree of a rational function $f(x)/g(x)$, assuming that this fraction is in lowest terms (i.e. $f(x)$ and $g(x)$ have no common polynomial factors), is the maximum of the degree of $f(x)$ and the degree of $g(x)$.
In your case, you have a function $r(x)$ where the most negative exponent is $x^{-N}$. As I will show below, this coincides with the definition of the degree of a rational function, i.e. the degree of $r(x)$, so the name for the most negative exponent in this case is just called the degree of the rational function.
To find the degree of the function $r(x)$ you made, we first need to put $r(x)$ into the form of one polynomial divided by another polynomial. To do this, we just need to factor out $1/x^N$ from the sum: $$r(x)=\frac{\sum_{i=n}^N a_i x^{N-i}}{x^N}$$ Assuming $a_N\neq 0$, this fraction is in lowest terms because the only factors of the denominator polynomial are powers of $x$, while the numerator polynomial has a constant term of $a_N\neq 0$ and thus does not have $x$ as a factor.
The degree of the numerator polynomial, $\sum_{i=n}^N a_i x^{N-i}$, is at most $N-n$ since the highest-degree term, if $a_n\neq 0$, is $x^{N-n}$. The degree of the denominator polynomial, $x^N$, is just $N$. Thus, the degree of $r(x)$ is the maximum of $N-n$ and $N$, which is just $N$, which is what we wanted to show.