Assume that the process stops when someone can't fit.

I believe the distribution of the amount of overshoot is known as a *ladder height distribution*, and that this is in Feller's classic text, but I don't have that handy.

Let $f(x)$ be the expected number of people up to and including the first who is turned away because there is no room. Let $\mu$ be the expected value of the distribution of widths. Let $m_2$ be the second moment which we'll assume is finite. If you rule out lattice distributions, so that the probability of $n$ people exactly fitting goes to $0$, then $$\begin{eqnarray} f(x) & = & \frac{x}{\mu} + \frac{m_2}{2\mu^2} + o(1) \newline & = & \frac{x}{\mu} + \frac{\sigma^2}{2\mu^2} + \frac{1}{2} + o(1). \end{eqnarray}$$

The number of people before the last is one lower, of course. I think when the second moment is infinite, the difference between $f(x)$ and $x/\mu$ is unbounded, and if the mean doesn't exist, then $f(x)$ is sublinear.

I posted a sketch of a proof assuming the distribution is continuous on a poker site. A more intuitive heuristic is to ask what happens if you choose the finish line randomly. How much does a width $w$ which occurs with probability $p$ contribute to the average overshoot? The chance that the finish line is within this width is $pw/\mu$, and when a width $w$ causes the overshoot, the average overshoot is $w/2$. So, the contribution to the average overshoot is $pw^2 / (2 \mu)$. Integrating over all widths gives an average overshoot of $\frac{m_2}{2\mu}$. The extra factor of $\mu$ in the denominator occurs because an overshoot of $y$ means you take an extra $y/\mu$ samples.

A related problem asks for the expected number of straws it takes to break a camel's back, if the straws have weights which are uniformly distributed on $[0,1]$, and the camel can hold a weight of $x$. If $x=1$, the probability that $n$ straws have weight less than $1$ is $1/n!$, so the expected number of straws is $\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}+... = e$. Asymptotically, it takes an average of about $2x+2/3 = \frac{x}{1/2} + \frac{1/3}{2*1/2^2}$.