Are you sure that it is an integer?

Note that $\displaystyle 7-4\sqrt{3} = 2^2-2\left(2\right)\left(\sqrt{3}\right)+\left(\sqrt{3}\right)^2=\left(2-\sqrt{3}\right)^2$ . Hence $\displaystyle \sqrt{7-4\sqrt{3}}=2-\sqrt{3}$ .

Now $\displaystyle 5\sqrt{7-4\sqrt{3}}+\sqrt{139+n\sqrt{3}}=10-5\sqrt{3} +\sqrt{139+n\sqrt{3}}$ is an integer if $\displaystyle \sqrt{139+n\sqrt{3}} = 5\sqrt{3} + p$ for some integer $p$ .

This means that $\displaystyle 139+n\sqrt{3} = \left(5\sqrt{3} + p\right)^2 = 75+p^2 +10p\sqrt{3}$ . Thus $\displaystyle 139 = 75+p^2$ and $\displaystyle n=10p$ . The former gives $\displaystyle p= \pm 8$ , and hence $\displaystyle (p,n)=(8,80), (-8,-80)$ .

Hence there are $\boxed{2}$ such $n$ .

Let the given be $N$ . Then

$\begin{aligned} N & = 5\sqrt{7-4\sqrt 3} + \sqrt {139+n\sqrt 3} \\ & = 5\sqrt{2^2-2(2)\sqrt 3 + (\sqrt 3)^2} + \sqrt{139+n\sqrt 3} \\ & = 5\sqrt{(2-\sqrt 3)^2} + \sqrt{139+n\sqrt 3} \\ & = 5(2-\sqrt 3) + \sqrt \blue{139+n\sqrt 3} & \small \blue{\text{Make }139+n\sqrt 3 = (m+5\sqrt 3)^2} \\ & = 10 - \red{5\sqrt 3} + \sqrt \blue{(m+\red{5\sqrt 3})^2} & \small \blue{\text{so that }\red{5\sqrt 3}\text{ disappears.}} \\ & = 10 - \red{\cancel{5\sqrt 3}} + m + \red{\cancel{5\sqrt 3}} \end{aligned}$

Then we have $139 + n\sqrt 3 = (m+5\sqrt 3)^2 = m^2 + 10m\sqrt 3 + 75$ .

$\implies \begin{cases} m^2 + 75 = 139 & \implies m^2 = 64 \implies m = \pm 8 \\ n = 10m = \pm 80 & \implies N = 10+m = 18, 2 \end{cases}$

Therefore there are $\boxed 2$ integer $n$ 's satisfying the equation.