16. (2005 Belarusian Mathematical Olympiad) Let $a, b$ be positive integers such that $79 \mid(a+77 b)$, and $77 \mid(a+79 b)$, find the smallest possible value of the sum $a+b$.
16. Notice that
$79|(a+77 b) \Leftrightarrow 79|(a-2 b) \Leftrightarrow 79|(39 a-78 b) \Leftrightarrow 79|(39 a+b)$,
$77|(a+79 b) \Leftrightarrow 77|(a+2 b) \Leftrightarrow 77|(39 a+78 b) \Leftrightarrow 77|(39 a+b)$,
thus, $79 \times 77 \mid(39 a+b)$,
therefore, $39 a+b=79 \times 77 k, k \in \mathbf{N}_{+}$.
Notice that
$$
39 a+39 b=79 \times 77 k+38 b=\left(78^{2}-1\right) k+38 b=\left(78^{2}-39\right) k+38(k+b),
$$

so, $39 \mid(b+k)$, hence $b+k \geqslant 39$.
Thus, $39 a+39 b \geqslant\left(78^{2}-39\right)+38 \times 39$, which means
$$
a+b \geqslant 156-1+38=193 \text {. }
$$

It is easy to see that $b=38, a=155$ satisfies the given conditions, therefore,
$$
a+b=193 \text {. }
$$

Hence, $(a+b)_{\min }=193$.