Define a sequence $\{a_{n}\}_{n=1}^{\infty}$ with $a_{1} = 1$ and $a_{n+1} = a_{n} + \lfloor \sqrt{a_{n}} \rfloor$ for all $n \geq 1$. Determine the smallest $N$ such that $a_{N} > 2017$.

Let's think step by step and output the final answer within \boxed{}.
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