1. Given that $a$ and $b$ are positive integers, and satisfy $\frac{a+b}{a^{2}+a b+b^{2}}=\frac{4}{49}$. Then the value of $a+b$ is $\qquad$ ـ.
II. 1. 16.
Let $a+b=4k$ ($k$ is a positive integer), then
$$
a^{2}+a b+b^{2}=49 k,
$$

which means $(a+b)^{2}-a b=49 k$.
Therefore, $a b=16 k^{2}-49 k$.
It is easy to see that $a, b$ are the two positive integer roots of the quadratic equation in $x$
$$
x^{2}-4 k x+\left(16 k^{2}-49 k\right)=0
$$

By $\Delta=16 k^{2}-4\left(16 k^{2}-49 k\right) \geqslant 0$, we get $0 \leqslant k \leqslant \frac{49}{12}$.
Since $k$ is a positive integer, thus, $k=1,2,3,4$.
It is easy to verify that when $k=1,2,3$, equation (1) has no positive integer roots; when $k=4$, equation (1) becomes $x^{2}-16 x+60=0$. Solving it, we get $x=10$ or $x=6$, which meets the requirements. Therefore, $a+b=4 k=16$.