Equivalent of Kolmogorov-Smirnov test for integer data? Is there an equivalent of the two-sample Kolmogorov-Smirnov test for integer data (not count data, as it can include negative integers)?
The Kolmogorov-Smirnov test does not perform well in the presence of lots of ties, which are obviously common with integers.
 A: The Permutation test could be applied here as well. The idea is as follows.
Let $X_1,...,X_m\sim F$ and $Y_1,...,Y_n\sim G$ be two independent samples and consider testing the hypothesis $H_0:F=G$ vs. $H_1:F\neq G$. For this purpose, label your data as follows
\begin{array}{c c}
1 & X_1\\
1 & X_2\\
\vdots & \vdots\\
1 & X_m\\
2 & Y_1\\
2 & Y_2\\
\vdots & \vdots\\
2 & Y_n\\
\end{array}
Now, let $T$ be an statistic of the sample $S=\{X_1,...,X_m,Y_1,...,Y_n\}$ and the labels $L=\{1,1,...,2,2,...,2\}$. 
If $H_0$ is true, then the labeling is superfluous. 
Now, permute the group labels and recalculate the test statistic a large number of times, say $B$. 

The one-sided p-value of this test is calculated as the proportion of sampled permutations where the difference in means was greater than or equal to $T(S,L)$. The two-sided p-value of the test is calculated as the proportion of sampled permutations where the absolute difference was greater than or equal to $\mbox{abs}(T(S,L))$. See

A toy example
Let $X_i \sim \text{Poisson}(10)$, $i=1,...,m=100$, and $Y_j \sim \text{Poisson}(11)$, $j=1,...,n=100$. Consider the statistic $T=\text{mean of Group 1} - \text{mean of Group 2}$. The permutation method using this statistic is implemented below.
rm(list=ls)
set.seed(1)
# Sample size
ns=100
#Simulated data
x = rpois(ns,11)
y = rpois(ns,10)

# Observed statistic    
T.obs = mean(x) - mean(y)

# Pooled data
SL = rbind(cbind(rep(1,ns),x),cbind(rep(2,ns),y))

# Resampling
B=10000
T = rep(0,B)

for(i in 1:B){
samp = sample(SL[,1])
ind1 = which(samp==1)
ind2 = which(samp==2)
T[i] = mean( SL[ind1,2] )- mean( SL[ind2,2] )
}

# p-value
p.value = length(which(abs(T)>abs(T.obs)))/B

I do not know how robust is this method, but after some experiments it seems to perform moderately well. Note that the choice of the statitic $T$ is open and therefore one must be careful on making a meaningful choice in the context of your problem as the performance depends on both the statistic and the sample size.
I hope this helps.
A: I would suggest the two sample chi square test where you bin the data and compare the binned total with an "expected number" that would fall within the binbased on the pooled sample.  This has a generalization to k greater than 2. I am assuming that you are not requiring another test of the emprical cdf form.  I think that entire class of test could have some trouble when there are a lot of ties.
Here is a reference that shows you precisely how the two-sample chi square test statistic is calculated along with the degrees of freedom for the asymptotic chi square distirbution.
