I could just start cutting them off

$\begin{aligned} a-b & =\frac{{1}^{2}}{1}+\frac{{2}^{2}-{1}^2}{3}+\frac{{3}^{2}-{2}^{2}}{5}+...+\frac{{1001}^{2}-{1000}^{2}}{2001}-\frac{{1001}^{2}}{2003} \\ & =1+1+1+...+1-\frac{{1001}^{2}}{2003} \\ & =1001-\frac{{1001}^{2}}{2003} \\ & =1001\left( 1-\frac { 1001 }{ 2003 } \right) \\ & =\frac { 1001\times 1002 }{ 2003 } \\ & = 500.749 \end{aligned}$

The closest integer of $500.749$ is $\huge \color{#3D99F6}{\boxed {501}}$

a = the sum from n=1 to 1001 of n^2/2n-1

b = the sum from n=1 to 1001 of n^2/2n+1

let a(n) = n^2/2n-1

and b(n) = n^2/2n+1

a(n+1) = (n+1)^2 / 2n+1

= n^2+2n+1 / 2n+1

= (n^2/2n+1) + 1

n^2/2n+1 = b(n), therefore:

a(n+1) - b(n) = 1

Therefore

(a-b) = 1^2/1 + 1000 - 1001^2/2003

(a-b) = 500.7498....