We have $$a_1 = \{ \sqrt{1} \} - \{ \sqrt{2} \} + \{ \sqrt{3} \} - \{ \sqrt{4} \} = (1-\sqrt{2}) + (\sqrt{3}-1) = \sqrt{3} - \sqrt{2}.$$ Therefore, it suffices to prove that $$\sqrt{3} - \sqrt{2} > 0.2.$$ Note that $(2.48)^2 = 6.1504 > 6,$ so $2.48 > \sqrt{6}.$ Therefore, it follows that $$5 - 2\sqrt{6} > 0.04.$$ Taking the square root of both sides yields $$\sqrt{3} - \sqrt{2} > 0.2,$$ so we are done. $\square$

Proof for >0: Obviously, there are an infinite number of natural numbers $n$ such that $$\lfloor \sqrt{n} \rfloor = \lfloor \sqrt{n+1} \rfloor = \lfloor \sqrt{n+2} \rfloor = \lfloor \sqrt{n+3}\rfloor.$$ In these cases, we have $$a_n = \{ \sqrt{n} \} - \{ \sqrt{n + 1} \} + \{ \sqrt{n + 2} \} - \{ \sqrt{n + 3} \}= \sqrt{n} - \sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}.$$ Note that $$n^2 + 3n < n^2 + 3n + 2,$$ for all $n \in \mathbb{Z}_{\geq 1}.$ Manipulating this yields $$\sqrt{n^2 + 3n} < \sqrt{n^2+3n+2}$$ $$-2\sqrt{n^2+3n} > -2\sqrt{n^2+3n+2}$$ $$n+3 - 2\sqrt{n^2+3n} + n > n+2 - 2\sqrt{n^2+3n+2} + n+1$$ $$(\sqrt{n+3} - \sqrt{n})^2 > (\sqrt{n+2} - \sqrt{n+1})^2$$ $$(\sqrt{n} - \sqrt{n+3})^2 > (\sqrt{n+1} - \sqrt{n+2})^2$$ $$\sqrt{n} - \sqrt{n+3} > \sqrt{n+1} - \sqrt{n+2}$$ $$\sqrt{n} - \sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}>0,$$ so we are done with this part. Proof for <0: Let $n$ be a perfect square such that none of the rest of the numbers are perfect squares. (Obviously, there is an infinite number of such constructions.) Then, we have $$a_n = -\sqrt{n+1} + \sqrt{n+2} - \sqrt{n+3}.$$ Clearly, $n+2 < n+3,$ so we have $$\sqrt{n+2} < \sqrt{n+3}$$ which means that $a_n$ must be negative, as desired. $\square$