The values in table I seem to be an average of the values in table II (rather than a direct computation with the quantile function).
The values in table II correspond very well with a normal distribution that has standard deviation $2000/7$. The upper boundaries will be equal to the percentile scores of the normal distribution rounded down and the lower boundaries will be equal to the percentile scores of the normal distribution rounded up. Below you see a comparison of three methods.
The logistic distribution or a normal distribution with standard deviation $200 \sqrt{2}$ have large differences.
For the normal distribution with standard deviation $2000/7$ there are some small discrepancies, but only of the order of $\pm 1$ and that can be due to round-off errors and use of tables with lower precision* or possibly a typo.
p = seq(0.505,0.995,0.01)
elo1 = qnorm(p,0,2000/7)
elo2 = qnorm(p,0,200*sqrt(2))
elo3 = -log10(1/p-1)*400
x = c(3,10,17,25,32,39,46,53,61,68,76,83,91,98,106,113,121,129,137,145,153,162,170,179,188,197,206,215,225,235,245,256,267,278,290,302,315,328,344,357,374,391,411,432,456,484,517,559,619,735)
plot(x,x-x, type = "l", xlab = "Elo's upper boundary", log = "", ylim = c(-2,15), ylab = "difference with table")
points(x,elo1-x, pch = 20)
points(x,elo2-x, pch = 2)
points(x,elo3-x, pch = 4)
legend(1,15, c("normal 2000/7", "normal 200*sqrt(2)", "logistic distribution"), cex = 1, pch = c(20,2,4))
lines(x,x*0+1)
In 'The rating of chess players, past and present' Elo writes an example
For example. let D = 160. Then z = 160/282.84 = .566. The table gives .7143 and .2857 as the areas of the two portions under the curve. These probabilities are rounded to two figures in table 2.11
This suggests that he used sufficiently accurate tables with at least 4 figures. An example could be Table X (copied from Pearson) in Garrett's 'statistics in psychology and education' which occurs as a reference.