Just a functional equation

$\begin{array} {rll} P(m,n): & f(m+n) = mn + f(m) + f(n) + 1 \\ P(0,0): & f(0) = 0 + f(0) + f(0) + 1 \\ & \color{#3D99F6} \implies f(0) = - 1 \\ P(-n,n): & {\color{#3D99F6}f(0)} = -n^2 + f(-n) + f(n) + 1 \\ & {\color{#3D99F6}-1} = -n^2 + f(-n) + f(n) + 1 \\ & \color{#3D99F6} \implies f(-n) + f(n) = n^2 - 2 \end{array}$

Now we have:

$\begin{aligned} S & = \sum_{n = -100}^{100} f(n) \\ & = \sum_{n = 0}^{100} f(-n) + \sum_{\color{#3D99F6}n = 1}^{100} f(n) \\ & = \sum_{\color{#D61F06}n = 0}^{100} \big(f(-n) + f(n)\big) \color{#D61F06} - f(0) \\ & = \sum_{n = 0}^{100} \left(n^2 - 2\right) + 1 \\ & = \frac {100(101)(201)}6 - 2(101) + 1 \\ & = \boxed{338149} \end{aligned}$

We have $f(0) = -1$ , and also,

$f(-n) = f(n-2n) \\ = -2n^2 + f(n) + f(-2n) + 1 \\ = -2n^2 + f(n) + [n^2 + 2f(-n) +1] + 1 \\ \iff f(n) + f(-n) = n^2 - 2.$

Therefore,

$\sum_{n=-100}^{100}{f(n)} = f(0) + \sum_{n=1}^{100}{f(n)+f(-n)} = -1 + \frac{100 \cdot 101 \cdot 201}{6} - 200 = \boxed{338149} \text{.}$

Bonus:

A substitution of $g(n) = f(n) - f(-n)$ shows that $g(n)$ satisfies

$\displaystyle g(m+n) = g(m) + g(n)$ ,

which is Cauchy's functional equation over the integers/rationals, and can be shown to have the unique family of solutions $g(n) = cn, c \in \mathbb{Z}$ . This, along with the result above, form simultaneous equations in $f(n)$ and $f(-n)$ , which can be solved to obtain

$f(n) = \boxed{\frac{n^2 + cn -2}{2}}; c,n \in \mathbb{Z}$ .

f(0)=f(n-n)=-n^2+f(-n) +f(n)+1 and f(n)=f(0+n)=0+f(0)+f(n)+1=f(0)+f(n)+1. Therefore f(0)=-1 and f(n)+f(-n) =n^2-2. This implies that f(n) is a 2nd. degree polynomial. Let f(n)=a(n^2)+bn+c. Then f(-n)=a(n^2)-bn+c, and so f(n)+f(-n)=2a(n^2)+2c.Therefore 2a=1 and 2c=-2 or a=1/2 and c=-1. This yields f(n)=1/2[n^2+2bn-2]. Therefore the given sum is ((1/2)(2)(100)(101)(201)(1/6))-201 or 338149