Let $$n$$ be an odd natural number such that the variance of 1, 2, 3, 4, ..., n is 14. Then $$n$$ is equal to _________.

We are given the consecutive natural numbers $$1,2,3,\dots ,n$$ and told that their variance equals $$14$$. Because every positive integer from $$1$$ to $$n$$ appears exactly once, the total number of observations is also $$n$$. Throughout this solution we shall regard the data set as the whole population, so we use the population variance formula

$$\sigma^{2}= \dfrac{1}{N}\sum_{i=1}^{N}(x_{i}-\mu)^{2},$$

where $$N$$ is the number of observations, $$x_{i}$$ denotes an individual observation and $$\mu$$ denotes the mean (average) of all the observations.

First we determine the mean. The sum of the first $$n$$ natural numbers is well known to be

$$\sum_{i=1}^{n} i=\dfrac{n(n+1)}{2}.$$

Dividing this sum by the total number of observations $$n$$ gives the mean

$$\mu=\dfrac{\dfrac{n(n+1)}{2}}{n}= \dfrac{n+1}{2}.$$

Next we require the second moment, that is, the mean of the squares. The sum of the squares of the first $$n$$ natural numbers is

$$\sum_{i=1}^{n} i^{2}= \dfrac{n(n+1)(2n+1)}{6}.$$

Hence the mean of the squares, often written as $$E[X^{2}]$$, equals

$$E[X^{2}]=\dfrac{1}{n}\sum_{i=1}^{n} i^{2}= \dfrac{1}{n}\cdot\dfrac{n(n+1)(2n+1)}{6}= \dfrac{(n+1)(2n+1)}{6}.$$

The variance can be computed by the identity

$$\sigma^{2}=E[X^{2}] - \mu^{2}.$$

Substituting the expressions we have just obtained, we write

$$\sigma^{2}= \dfrac{(n+1)(2n+1)}{6} - \left(\dfrac{n+1}{2}\right)^{2}.$$

Because it is given that the variance equals $$14$$, we set the above expression equal to $$14$$:

$$\dfrac{(n+1)(2n+1)}{6} - \left(\dfrac{n+1}{2}\right)^{2}=14.$$

To simplify, notice that each term contains the factor $$(n+1).$$ We write

$$\dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^{2}}{4}=14.$$

Factorising $$(n+1)$$ from both numerators gives

$$(n+1)\left[\dfrac{2n+1}{6}-\dfrac{n+1}{4}\right]=14.$$

We now combine the expressions in the square brackets by taking a common denominator of $$12$$:

$$\dfrac{2n+1}{6}= \dfrac{4n+2}{12},\qquad \dfrac{n+1}{4}= \dfrac{3n+3}{12}.$$

Subtracting inside the brackets, we obtain

$$\dfrac{4n+2}{12}-\dfrac{3n+3}{12}= \dfrac{4n+2-3n-3}{12}= \dfrac{n-1}{12}.$$

Therefore

$$(n+1)\left(\dfrac{n-1}{12}\right)=14.$$

Multiplying both sides by $$12$$ to clear the denominator, we have

$$(n+1)(n-1)=14\times12=168.$$

Recognising the left side as the difference of squares, we write

$$n^{2}-1=168.$$

Adding $$1$$ to both sides produces

$$n^{2}=169.$$

Taking the positive square root (because $$n$$ is a natural number) yields

$$n=13.$$

Finally, the question stipulates that $$n$$ must be odd; $$13$$ is indeed odd, so our solution is fully consistent.

So, the answer is $$13$$.