A problem by Josh Rowley

It follows by Cauchy-Schwarz Inequality that $(\sum_{i=1}^{2014}a_i)^2\le(\sum_{i=1}^{2014}\frac{a_i^2}{i})(\sum_{i=1}^{2014}i)$ .

Hence the minimum value possible is $\frac{100^2}{\frac{2014*2015}{2}} \cong 0.00492828$ and therefore the answer is $\boxed{49}$ .

From Titu's Lemma, we know that $\sum_{i=1}^{2014} \dfrac{a_{i}^2}{b_{i}} \ge \dfrac{( a_{1} + a_{2} + ... + a_{2014} )^2}{ b_{1} + b_{2} + ... + b_{2014} }$ . So, if we let $b_{i} = i$ then we have that the minimum is $\dfrac{100^2}{1007 \times 2015}$ . Then multiplying by 10000 and applying the floor function we achieve an answer of $49$