16. There exists a positive integer $n$, such that both $8 n-7$ and $18 n-35$ are perfect squares. Then all positive integers $n=$. $\qquad$ _.
【Analysis】Let's assume $\left\{\begin{array}{l}18 n-35=a^{2} \\ 8 n-7=b^{2}\end{array} \Rightarrow\left\{\begin{array}{l}72 n-140=4 a^{2} \\ 72 n-63=9 b^{2}\end{array} \Rightarrow(3 b)^{2}-(2 a)^{2}=77\right.\right.$ Therefore, $(3 b-2 a)(3 b+2 a)=77=1 \times 77=7 \times 11$
Solving, we get $\left\{\begin{array}{l}a=19 \\ b=13 \\ n=22\end{array}\right.$ or $\left\{\begin{array}{l}a=1 \\ b=3 \\ n=2\end{array}\right.$