How do I sample multiple normally distributed items and maintain a fixed sum? Suppose jars are being filled with Red, Green and Blue balls in an approximately fixed ratio. The total number of balls that fits in a jar is fixed (say N=1000). I know the mean and standard deviation of the ratios; let's say means for (R,G,B) are 0.5,0.4,0.1 and stdevs are 0.05,0.05,0.02. They appear to be approximately normally distributed. 
How can I construct samples from the distributions of R+G+B balls such that they sum to N?
If I sample R and G independently and calculate B = N-R-G, then G may not follow the original distribution and may even be negative. If I sample independently from R, G and B, the total will not be N; if I renormalise the sample after, they will no longer follow the original distribution. 
Any thoughts welcome.
 A: Those distributions are discrete, and bounded between 0 and N inclusive; they are only approximately normal, and in this case many of your problems are caused by assuming they're normal when they aren't.
When you say you know the mean and sd of the ratios, are these estimates of populations proportions with associated standard errors, or are these actual population proportions that can vary?
In the second case, I'd suggest sampling a multinomial distribution with the required proportions, perhaps selecting the varying the proportions using a Dirichlet. In the first case it would be a little more complex, but you might use a Dirichlet-multinomial distribution directly to incorporate the uncertainty.
What are you working in? Do you have something that will simulate a multinomial? Or a binomial? Something that will simulate a beta or a gamma?
Here's an example just taking the proportions as fixed at some known value (perhaps itself generated from appropriate distributions related to the given mean and sds) and assuming you can generate binomially distributed random numbers:
Assume $(p_R,p_G,p_B)$ are given (and sum to 1).
1) generate $R\sim \operatorname{Bin}(1000,p_R)$
2) generate $G\sim \operatorname{Bin}(1000-R,\frac{p_G}{p_G+p_B})$
3) let $B = 1000 - R - G$ 
If $(p_R,p_G,p_B)=(0.5,0.4,0.1)$ then that's $R\sim \operatorname{Bin}(1000,0.5)$, $G\sim \operatorname{Bin}(1000-R,0.8)$ and the rest are $B$'s.
A more complete answer will require better information about what those 'standard deviations' represent.
