we have$ a_{i} + a_{i - 2}\geq 2a_{i - 1} + 1$ so we have $ a_{i} - a_{i - 1}\geq a_{i - 1} - a_{i - 2} + 1\geq.....a_{2} - a_{1} + (i - 2)\geq i - 1$ so we have $ a_{i}\geq \frac{i^2 - i + 2}{2}$. so min of $ a_{i}$ is $ \frac {i^2 - i + 2}{2}$ .and its easy to check the condition because $ i^2 + l^2 > j^2 + k^2$ .so the min of$ a_{2008}$ is 2015029.

Note: It says in the question that $ 1 \leq i < j \leq k < l$ and not $ 1 \leq i < j < k < l$. The answer to the problem is $ 2^{2007}$.

I may be wrong but i think that the sequence a(n) = n*(n+1)/2 works. So a(2008) = 1004*2009 = 2017036

sumita is correct.

I think 673015