1. Given $a, b$ are positive integers, and satisfy $\frac{a+b}{a^{2}}=\frac{4}{40}$. Then the value of $a+b$ is $\qquad$
Let $a+b=4k$ ($k$ is a positive integer), then $a^{2}+ab+b^{2}:=49k$, i.e., $(a+b)^{2}-ab=49k$.
$$
\therefore ab=16k^{2}-49k \text{. }
$$

It is easy to know that $a, b$ are the two positive integer roots of the equation about $x$
$$
x^{2}-4kx+(16k^{2}-49k)=0
$$

By $\Delta=16k^{2}-4(16k^{2}-49k) \geqslant 0$, we get $0 \leqslant k \leqslant \frac{49}{12}$.
$\because k$ is a positive integer, $\therefore k=1,2,3,4$.
It is easy to verify that when $k=1,2,3$, equation (1) has no positive integer roots; when $k=4$, equation (1) is $x^{2}-16x+60=0$, the solutions are $x=10$ or $x=6$. Therefore, $a+b=4k=16$.