smc

If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x<1$, then $x$ is odd, and $t_{(x-1)}=\frac{1}{t_x}$. Otherwise, you keep on subtracting 1 and halving x until $t_{\frac{x}{2^n}}<1$. We can use this logic to go backwards until we reach $t_1=1$, like so: $$\begin{gathered} t_n=\frac{19}{87} \\ {t}_{n-1}=\frac{87}{19} \\ {t}_{\frac{n-1}{2}}=\frac{68}{19} \\ {t}_{\frac{n-1}{4}}=\frac{49}{19} \\ {t}_{\frac{n-1}{8}}=\frac{30}{19} \\ {t}_{\frac{n-1}{16}}=\frac{11}{19} \\ {t}_{\frac{n-1}{16}-1}=\frac{19}{11} \\ {t}_{\frac{\frac{n-1}{16}-1}{2}}=\frac{8}{11} \\ {t}_{\frac{\frac{n-1}{16}-1}{2}-1}=\frac{11}{8} \\ {t}_{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}}=\frac{3}{8} \\ {t}_{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}=\frac{8}{3} \\ {t}_{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{2}}=\frac{5}{3} \\ {t}_{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}}=\frac{2}{3} \\ {t}_{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}-1}=\frac{3}{2} \\ {t}_{\frac{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}-1}{2}}=\frac{1}{2} \\ {t}_{\frac{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}-1}{2}-1}=2 \\ {t}_{\frac{\frac{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}-1}{2}-1}{2}}=t_1=1\Rightarrow \frac{\frac{\frac{\frac{\frac{\frac{n-1}{16}-1}{2}-1}{2}-1}{4}-1}{2}-1}{2}=1\Rightarrow n=1905 \end{gathered}$$, so the answer is $\boxed{15}$.