Conditional distribution of $(X_i)_i$ given $\sum\limits_i X_i$ when $(X_i)_i$ is i.i.d. 
Suppose $X_1,X_2,\ldots, X_n$ are i.i.d. random variables. Is there some way to determine the distribution of $(X_1,\ldots,X_n)$ given $S_n := X_1+\ldots+X_n$? 

In the discrete case it is easy just using the definition of conditional expected value, but what about in the continuous case? 
 A: Expanding on my first comment... 
General formula
Let $S$ be a random variable with PDF $f_S(s)$ for $s \in \mathbb{R}$. Let $A$ be an event with $P[A>0]$. Then:
$$ P[A|S=s] = \frac{f_{S|A}(s|A)P[A]}{f_S(s)} $$ 
For intuition about this formula, it is easy to verify that for any interval $[a,b]$: 
\begin{align}
\int_{s=-\infty}^{\infty} P[A|S=s]f_S(s)ds &= P[A] \\
\int_{s \in [a,b]} P[A|S=s] f_S(s) &= P[A \cap \{ S \in [a,b]\}] 
\end{align} 
Application to your problem
Let $\{X_i\}_{i=1}^n$ be i.i.d., let $S=\sum_{i=1}^n X_i$.  We want to find $P[(X_1, \ldots, X_n) \leq (x_1, ..., x_n) | S=s]$ for all relevant values of $s \in \mathbb{R}$.  For notational simplicity define $X=(X_1, ..., X_n)$, $x=(x_1, ..., x_n)$, $A_x = \{X\leq x\}$.  Note that:
$$P[A_x] = P[X\leq x] = P[X_1\leq x_1]\cdots P[X_n\leq x_n] $$
 Assume $S$ has PDF $f_S(s)$. We want to compute $P[A_x|S=s]$.  Applying the above formula gives: 
\begin{align*}
P[A_x | S=s] &= \frac{f_{S|A_x}(s|A_x)P[A_x]}{f_S(s)} \\
&= \frac{f_{S|A_x}(s|A_x)P[X_1\leq x_1]\cdots P[X_n\leq x_n]}{f_S(s)}
\end{align*} 
You can find $f_S(s)$ by $n$-fold convolution of the PDFs of $f_X(s)$ (assuming such exist).  You can find $f_{S|A_x}(s|A_x)$ by: 
\begin{align}
f_{S|A_x}(s|A_x) &= \frac{d}{ds} P[S \leq s| A_x] \\
&=  \frac{1}{P[A_x]}\frac{d}{ds}P[\{S\leq s \} \cap A_x]\\
&= \frac{\frac{d}{ds} P[X_1\leq x_1, ..., X_n \leq x_n, X_1+...+X_n\leq s]}{P[X_1\leq x_1]\cdots P[X_n\leq x_n]}
\end{align}
Computing probabilities with the conditional CDF
Notice that the above gives the conditional cumulative distribution function  (CDF) rather than the conditional PDF.  Conditional PDFs given  $X_1 + ... +X_n=s$ are hard to define since, as you note, this restricts $(X_1, ..., X_n)$ to a multidimensional set of measure zero in $\mathbb{R}^n$.  So real-valued functions defined over that measure-zero set would integrate to 0, not to 1. One would need to use PDFs with multidimensional impulses, which are tricky.  However, there is no need to use impulses:  Working with the conditional CDF gives all you need.  
For example, suppose $n=2$ and let $B$ be the line segment in $\mathbb{R}^2$ between the points $(0,1)$ and $(1,0)$.  How do we compute $P[(X_1, X_2) \in B | S=1]$?  We just observe: 
$$P[(X_1, X_2) \in B | S=1] = P[\underbrace{X_1 \leq 1, X_2 \leq 1}_{A_{(1,1)}} | S=1] $$
A 2-d picture to illustrate the above equation would be great, but I do not know how to post a picture on stackexchange. 
