When $n$ standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $S$. The smallest possible value of $S$ is
$\textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341$
Let $d_i$ be the number on the $i$th die.  There is a symmetry where we can replace each die's number with $d_i' = 7-d_i$.  Note that applying the symmetry twice we get back to where we started since $7-(7-d_i)=d_i$, so this symmetry is its own inverse.  Under this symmetry the sum $R=\sum_{i=1}^n d_i$ is replaced by $S = \sum_{i=1}^n 7-d_i = 7n - R$.  As a result of this symmetry the sum $R$ the sum $S$ have the same probability because any combination of $d_i$ which sum to $R$ can be replaced with $d_i'$ which sum to $S$, and conversely.  In other words, there is a one-to-one mapping between the combinations of dice which sum to $R$ and the combinations which sum to $S$.
When $R=1994$ we seek the smallest number $S=7n-1994$, which clearly happens when $n$ is smallest.  Therefore we want to find the smallest $n$ which gives non-zero probability of obtaining $R=1994$.  This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in $R=1994$ being impossible.  Thus $n = \left\lceil \frac{1994}{6} \right\rceil = 333$ and $S = 333\cdot 7 - 1994 = 337$.  The answer is $\textbf{(C)}$.