The value of V does not mean the number of positive scores, but the sum of these positive scores.
As well there is a measurement for the sum for the negative scores, that this test does not provide. A brief script for calculating the sum for positive and for negative scores is provided in the following example:
a <- c(214, 159, 169, 202, 103, 119, 200, 109, 132, 142, 194, 104, 219, 119, 234)
b <- c(159, 135, 141, 101, 102, 168, 62, 167, 174, 159, 66, 118, 181, 171, 112)
diff <- c(a - b) #calculating the vector containing the differences
diff <- diff[ diff!=0 ] #delete all differences equal to zero
diff.rank <- rank(abs(diff)) #check the ranks of the differences, taken in absolute
diff.rank.sign <- diff.rank * sign(diff) #check the sign to the ranks, recalling the signs of the values of the differences
ranks.pos <- sum(diff.rank.sign[diff.rank.sign > 0]) #calculating the sum of ranks assigned to the differences as a positive, ie greater than zero
ranks.neg <- -sum(diff.rank.sign[diff.rank.sign < 0]) #calculating the sum of ranks assigned to the differences as a negative, ie less than zero
ranks.pos #it is the value V of the wilcoxon signed rank test
[1] 80
ranks.neg
[1] 40
CREDITS: https://www.r-bloggers.com/wilcoxon-signed-rank-test/
(They also provide a nice context for it.)
You can compare also both of these numbers to their average (in this case, 60), that would be the expected value for each side, i.e. positive ranks summing 60 and negative ranks summing 60 means complete equivalence of the sides. Do positive ranks summing 80 and negative ranks summing 40 can also be considered equivalent? (i.e. could we just attribute this difference of "20" to stochastic reasons or is this distant enough for us to reject the hypothesis of no-difference?)
So, as they explain, the critical interval for this case is [25,95]. Checking on a table for critical values for the Wilcoxon rank signed test, the critical value for this example is 25 (15 pairs at 5% on a two-tailed test; and 120-25 = 95...). Meaning that the interval [40,80] is not "big enough" to discard the possibility that the differences are purely due to random sampling. (Consistently, the p-value is above the alpha).
To compare the sum of positive scores to the sum of negative scores helps to determine the significance of the difference, it enriches the analysis. Also, the positive ranks themselves are input for the calculation of the p-value of the test, therefore the interest in them.
But to extract meaning from a simply reported sum of positive ranks (V), only, I think that is not straightforward. In terms of providing information, I believe that the least to do is to also check the sum of the negative ranks, too, to have a more consistent idea of what is happening. (of course, along with general info, like sample size, p-value, etc).