Define $ g_i$ as the amount of goodbye received by friend $ i$, WLOG $ g_1 < g_2 < \cdots < g_{13}$ And $ k$ the number of times of Frederiks salutes. $ 10k=\sum_{i=1}^{13}{g_i}$ $ g_{13} \le k$ then $ 10k \le (g_{13}-12)+(g_{13}-11)+\cdots+(g_{13}-1)+g_{13}$ $ \le (k-12)+(k-11)+\cdots+(k-1)+k$ $ 0 \le 13k-78-10k$ $ 26 \le k$ and for the case $ k=26$ it is not difficult to find an example.

mszew wrote: ...and for the case $ k=26$ it is not difficult to find an example. That's what happens when you don't sit down and actually write down an example. Your argument until here is correct but not sharp. In fact only $k \ge 33$ is possible (and one can actually construct an example for $k=33$!). The point is that you forgot about the special rôle of the last friend to whom he only said goodbye in the last night i.e. only once! So you actually get $g_1=1$ and hence \[10k \le k+(k-1)+(k-2)+\dots+(k-11)+1=12k-65\]so that $k \ge 33$.

easy problem