What is the largest integer $n$ for which

$$
\frac{\sqrt{7}+2 \sqrt{n}}{2 \sqrt{7}-\sqrt{n}}
$$

is an integer?
If $n=28$, then the fraction is undefined. If $n \neq 28$, then denoting the value of the fraction by $A$, after rationalizing, we get

$$
A=\frac{14+2 n+5 \sqrt{7 n}}{28-n}
$$

If $A$ is an integer, then since the denominator is an integer, the numerator must also be an integer. Since $(14+2 n)$ is an integer, $5 \sqrt{7 n}$ must also be an integer, which means $\sqrt{7 n}$ is rational. Since $7 n$ is an integer, this is only possible if $7 n$ is a perfect square, i.e., $n$ is seven times a perfect square, $n=7 k^{2}$, and here we can assume that $k \geqq 0$. Substituting this into (1):

$$
A=\frac{14+14 k^{2}+35 k}{28-7 k^{2}}=\frac{(2 k+1)(k+2)}{(2-k)(2+k)}=\frac{2 k+1}{2-k}=-2-\frac{5}{k-2}
$$

The value of $A$ is an integer precisely when $k-2$ - which is an integer - is a divisor of 5. The largest absolute value of such a $k$ is 7, in which case $n=7^{3}=343$ and then $A=-3$.

Therefore, the largest integer $n$ for which the given expression is an integer is 343.