2. For an arithmetic sequence $\left\{a_{n}\right\}$ with a common difference of $d$ and all terms being positive integers, if $a_{1}=1919, a_{m}=1949, a_{n}=2019$, then the minimum value of the positive integer $m+n$ is . $\qquad$
$d=\frac{30}{m-1}=\frac{100}{n-1}=\frac{70}{n-m}$, when $d$ is the greatest common divisor of $30,100,70$, which is 10, $m+n$ is minimized. So $m=4, n=11 \Rightarrow m+n=15$.