Is a chi-squared test possible without the number of participants? I'm trying to work out whether it is possible to perform a chi-squared test or any other statistical test on a set of data without knowing the number of participants (e.g. 300), but instead knowing the percentages of their distributions. The question is specifically 4 groups with 18% in A, 15% in B, 19% in C and 48% in D. Is it possible to work with only this information?
 A: The chi-squared test requires the counts because those determine how uncertain each of the proportions is.    However, it's still possible to make some progress.
For instance, if the value of 19% represents 190 out of 1000 observed in a random sample, its standard error is only 1.24%; but if it represents 3 out of 16 observed, its standard error is 9.8%.  Thus, the very meaning of a proportion like "19%" depends on the count on which it is based (as well as on the sampling procedure, which here is assumed to be simple random sampling with replacement).
However, some information about the sample size can be gleaned from the data.  The fact that nearly-equal values of 18% and 19% appear implies the sample size is large enough to distinguish between 18% and 19%.  However, because the proportions are given only to two significant figures, we have to understand these percentages as being rounded values.  In the most extreme case, there's as much as a 2% difference because 18% could be 17.5% and 19% could be 19.5%.  Nevertheless, a sample size of at least 1/(2%) = 50 is needed to make that distinction.
We could therefore conservatively run a chi-squared test by converting the proportions to counts, assuming the smallest possible sample size consistent with the counts.  In this case, the proportions of 18%, 15%, 19%, and 48% translate to counts of 11, 9, 12, and 30, respectively (summing to 62).  The chi-squared test of equal proportions has a p-value of 0.04%.  If you find this small enough to be significant (and most people would), then you can be sure the test would be significant when applied to any larger sample size, since larger samples only make it easier to detect small differences.
If, on the other hand, your data were (say) 18%, 15%, 21%, and 46%, then we could only conclude the sample size was at least 33, rather than 50, with corresponding counts 6,5,7, and 15.  The chi-squared test statistic of 5.49% would usually not be considered significant.

Here is R code that estimates the smallest possible sample size for an array of proportions p, prints out the corresponding counts k, and performs the chi-squared test.  It requires you to specify the number of decimal places of precision (dp) used to represent the proportions.
p <- c(A=18, B=15, C=19, D=48)/100
dp <- 2
#
# Brute force search for smallest `n`.
#
for (n in 1:(10^dp)) {
  q <- round((k <- round(p*n)) / n, dp)
  if (isTRUE(all.equal(q, p))) break
}
print(k)


 A  B  C  D 
11  9 12 30


chisq.test(k)


Chi-squared test for given probabilities

data:  k
X-squared = 18.387, df = 3, p-value = 0.000366


