Yes, this is certainly possible. Let $d$ be the observed Cohen's d value based on a sample size of $n_1$ and $n_2$ individuals in the two groups. Then an approximate 95% confidence interval is often computed with $$d \pm 1.96 \sqrt{Var[d]},$$ where $$Var[d] = \frac{1}{n_1} + \frac{1}{n_2} + \frac{d^2}{2(n_1 + n_2)}.$$ Let $LB$ and $UB$ denote the lower and upper confidence interval bounds obtained in this manner.
Now if you are willing to assume that $n_1 = n_2 = n$, then $$Var[d] = \frac{8 + d^2}{4n}.$$ This implies that $$n = \frac{8 + d^2}{4 Var[d]}.$$ And since $$(UB - LB)^2 / (2\times1.96)^2 = Var[d],$$ we get $$n = \frac{(8 + d^2)(2 \times 1.96)^2}{4(UB - LB)^2}.$$
Let's try this out with R, using the metafor
package for the computation of the CI.
library(metafor)
n <- 25 ### sample size of group 1 and group 2
dat <- escalc("SMD", m1i=10.8, sd1i=2.5, n1i=n, m2i=9.3, sd2i=2.2, n2i=n)
dat <- summary(dat)
dat
This yields:
yi vi sei zi ci.lb ci.ub
1 0.6270 0.0839 0.2897 2.1642 0.0592 1.1948
Here, yi
is the d-value (with bias correction) and ci.lb
and ci.ub
are the confidence interval bounds. Suppose this is all that we know. Then:
c(with(dat, (8 + yi^2) * (2 * 1.96)^2 / (4 * (ci.ub - ci.lb)^2)))
yields:
[1] 25.00092
The slight discrepancy is due to using 1.96 and not qnorm(.975)
for the exact 97.5th quantile of a standard normal distribution. But in practice, the d value and CI bounds will only be given to a few decimals, so the sample size obtained in this manner won't be exact anyway. It should be fairly close though. For example, if results are reported to two decimals, then:
c(with(dat, (8 + round(yi,2)^2) * (2 * qnorm(.975))^2 / (4 * (round(ci.ub,2) - round(ci.lb,2))^2)))
yields:
[1] 25.26145
So, just round the value obtained.