Germany TST

Solution:

If 2013 is reliable, then there exists a positive integer index $i$ with $i=2013 a_{i} \geqslant 20 a_{i}$. In particular, the set $S$ of all positive integers $s$ for which $s \geqslant 20 a_{s}$ holds is non-empty. Thus $S$ contains a smallest element $j$, and this satisfies on the one hand
$$j \geqslant 20 a_{j}$$
and on the other hand $(j-1) \notin S$. The latter means that either $j=1$ must hold, or $j>1$ and $j-1<20 a_{j-1}$. However, because of $a_{j} \geqslant 1$, it follows from (2) that $j \geqslant 20$, which rules out the first of these two alternatives; hence it must be that $j \leqslant 20 a_{j-1}$, and combining this with (2) we obtain the chain of inequalities
$$j \geqslant 20 a_{j} \geqslant 20 a_{j-1} \geqslant j .$$
This can only hold if equality holds throughout. Because $j=20 a_{j}$, the index $j$ witnesses that 20 is, as claimed, reliable.