Here's an approximate answer: let $U_1,U_2,\ldots,U_n$ be the i.i.d. uniform random variables. I will consider them ordered from smallest to largest because it makes the arithmetic cleaner. We're then asking adding from smallest to largest, how many do we need to sum in order that $U_{i_1}+\ldots +U_{i_k}>\frac S2$, where $S=U_1+\ldots+U_n$.

Define additional random variables:
$$
\begin{align*}
X&=\#\{i\colon U_i<1/\sqrt 2\};\\
Y&=\sum_{U_i<1/\sqrt 2} U_i-n/4;\\
Z&=\sum_i U_i-n/2.
\end{align*}
$$
It's not hard to check that $\mathbb EX=n/\sqrt 2$, $\mathbb EY=\mathbb EZ=0$.

The ideal situation is when $Z=2Y$; that is the terms below $1/\sqrt 2$ have exactly the same sum as those above. To compensate for the difference, we need to move terms totalling $\frac Z2-Y$ into the "small number subset" (if $Z>2Y$)
or move terms totalling $Y-\frac Z2$ into the large number subset (if $Z<2Y$).

Since the terms on the boundary are of size approximately $\frac 1{\sqrt 2}$, one expects that the number such that $U_{i_1}+\ldots+U_{i_k}=S/2$ is well-approximated by $M=X+\sqrt 2(\frac Z2-Y)$.

From the definition, one sees
$$
M=\sum_{i=1}^n \Big( \mathbf 1_{U_i<1/\sqrt 2} + (1/\sqrt 2)U_i - \sqrt 2U_i\mathbf 1_{U_i<1/\sqrt 2}\Big).
$$
A calculation shows that $\mathbb EM=n/\sqrt 2$ and $\text{Var}(M)=n(9\sqrt 2-2)/12$. I anticipate that the expectation is correct to $O(1)$ and the variance is correct to $O(\sqrt n)$.

BTW: If anyone can tell me how make this rigorous, I'd very much like to know.

order statistics. $\endgroup$ – Anthony Quas Mar 14 '16 at 22:16