jmc

We will consider two cases.

Case 1: $x-3$ is nonnegative. In this case, we have $|x-3|=x-3$. Also, if $x-3$ is nonnegative, then $3-x$ (which is $-1$ times $x-3$) is nonpositive, which means that $|3-x|=-(3-x)$. So, substituting for the absolute values in the original equation, we have $$x-3-(3-x)-1=3.$$Solving this equation gives $x=5$.

Case 2: $x-3$ is negative. In this case, we have $|x-3|=-(x-3)$. Also, when $x-3$ is negative, then $3-x$ is positive, so $|3-x|=3-x$. So, substituting for the absolute values in the original equation, we have $$-(x-3)+3-x-1=3.$$Solving this equation gives $x=1$.

Combining our cases, the sum of the values of $x$ that satisfy the equation is $\boxed{6}$.

Notice that we could have solved this equation even faster by noticing that $|3-x|=|(-1)(x-3)|=|(-1)||x-3|=|x-3|$, so the original equation simplifies to $2|x-3|-1=3$, which gives us $|x-3|=2$. From here, we see that $x$ is 2 away from 3 on the number line, so $x$ is 5 or 1.