14. Note that $a_n, b_n >0$ for all $n\in \mathbb N$.
Without any loss of generality, we will assume from now on that $a_1\ge b_1$.
By induction, we can show that $a_n\ge b_n$ for all $n\in \mathbb N$.
Therefore, $a_{n+1}=a_n+\frac{1}{b_n}\ge a_n+\frac{1}{a_n}$ for all $n\in \mathbb N$.
In particular, $a_2\ge 2$ because the minimum of $x+\frac1x$ for positive real $x$ is 2.
Let $c_n:=a_n^2$. Hence, $$ c_{n+1} = a_{n+1}^2 \ge \left( a_n+ \frac{1}{a_n} \right)^2 = c_n + \frac{1}{c_n} +2 > c_n+2. $$ Since $c_2 = a_2^2 \ge 4$, we can prove by induction that $c_n>2n$ for $n\ge3$.
If $n\ge 1000^2/2 =500000$, then $c_n>1000^2$, and we have $$ a_n+b_n >a_n= \sqrt{c_n}>1000 $$ as required.
