Given,

The mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively,

So, mean will be,

 $\overline{x}=\frac{x_1+x_2+\dots \dots +9+10+\dots ..+x_{12}}{12}=\frac{9}{2}$

$\Rightarrow x_1+x_2\dots \dots +x_{12}+19=54$

Now removing the observation $9 \& 10$ and adding the observation $7 \& 14$,

So, the new total sum of the observation will be,

$x_1+x_2+\dots \dots +x_{12}+7+14=54-19+7+14$

Now new mean will be,

$\frac{x_1+x_2+\dots .+x_{12}+7+14}{12}=\frac{56}{12}$

$\Rightarrow {\overline{x}}_{\text{new}}=\frac{14}{3}$
Now using the formula of the variance we get,

$\frac{x_1^2+x_2^2+\dots \dots +x_{12}^2+9^2+{10}^2}{12}-{\left(\frac{9}{2}\right)}^2=4$

$\Rightarrow x_1^2+x_2^2+\dots \dots +x_{12}^2+9^2+{10}^2-81\times 3=4\times 12$

$\Rightarrow x_1^2+x_2^2+\dots \dots +x_{12}^2+9^2+{10}^2=291$

Now removing $9^2 \& {10}^2$ and adding $7^2 \& {14}^2$ we get,

$\Rightarrow x_1^2+x_2^2+\dots \dots +x_{12}^2+7^2+{14}^2=355$
New variance $=\frac{\sum x_i^2}{N}-(\overline{x}{)}^2$
$=\frac{355}{12}-{\left(\frac{14}{3}\right)}^2$
$=\frac{281}{36}$

Now on comparing with $\frac{m}{n}=\frac{281}{36}$ we get,

$\Rightarrow m=281$ and $n=36$

$\therefore m+n=317$