The equation $x-\frac{7}{x-3}=3-\frac{7}{x-3}$ has:
$\textbf{(A)}\ \text{infinitely many integral roots}\qquad\textbf{(B)}\ \text{no root}\qquad\textbf{(C)}\ \text{one integral root}\qquad$
$\textbf{(D)}\ \text{two equal integral roots}\qquad\textbf{(E)}\ \text{two equal non-integral roots}$
Both terms have a $-\frac{7}{x-3}$ term, so add $\frac{7}{x-3}$ to both sides.  This results in $x = 3$.
However, note that if $3$ is plugged back into the original equation, it results in
\[3-\frac{7}{0}=3-\frac{7}{0}\]
Since dividing by zero is undefined, $3$ is an extraneous solution.  That means there are no solutions, so the answer is $\boxed{\textbf{(B)}}$.