There has to be some trick to this!

The generalization of this problem in summation notation is $\sum_{k=1}^{n} \frac{k^3+1}{k+1}$ which is equal to $\sum_{k=1}^{n} (k^2-k+1)=\sum_{k=1}^{n}k^2-\sum_{k=1}^{n}k+\sum_{k=1}^{n}1=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n=n\left( \frac{(2n^2+3n+1)-(3n+3)+6}{6} \right)=n\left( \frac{n^2+2}{3} \right)$ with the use of appropriate summation formulas and simplifying. The sum in the problem is obtained by taking $n=2005$ , giving the value $2005 \left( \frac{2005^2+2}{3} \right)$ This number is odd, so it is not divisible by 2. It is not divisible by 3, since $2005 \left( \frac{2005^2+2}{2} \right) \equiv 1 \left( \frac{1^2+2}{3} \right)=1 \pmod 3$ Obviously, it is divisible by 5, because $5 \mid 2005$ (and 5 is the lowest prime factor of the sum).

It is not divisible by 7, since $2005 \left( \frac{2005^2+2}{3} \right) \equiv 3 \left( \frac{3^2+2}{3} \right) = 11 \equiv 4 \pmod 7$ It is divisible by 11, since $2005 \left( \frac{2005^2+2}{3} \right) \equiv 3 \left( \frac{3^2+2}{3} \right)=11 \equiv 0 \pmod {11}$ Since 5 is the lowest prime factor of the sum, then $\boxed{11}$ is the second smallest one.

A code could also be used to solve this (I'll just post this since there's already a solution posted for doing it manually) :)

imgur

This gives $\boxed{11}$ .