$\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2 n+1}{3}$ For $n=1,2,3, \ldots . ., 1000$ Value of $\frac{2 n+1}{3}=\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ respectively. Out of $\frac{3}{3}, \frac{5}{3}, \frac{7}{3}, \ldots \ldots \ldots, \frac{2001}{3}$ only first term $\left(\frac{3}{3}=1\right)$, fourth term $\left(\frac{9}{3}=3\right), 667^{\text {th }}$ term $\left(\frac{2001}{3}=667\right)$ are integers.
Hence, out of 1000 values of $\frac{2 n+1}{3}$, total number of integral values of $\frac{2 n+1}{3}$ $=333+1=334$
$\therefore \quad$ Required probability $=\frac{334}{1000}=0.334$