If the mean deviation about the mean of the numbers $$1, 2, 3, \ldots, n$$, where $$n$$ is odd, is $$\frac{5(n+1)}{n}$$, then $$n$$ is equal to ______.

We need to find $$n$$ (odd) such that the mean deviation about the mean of $$1, 2, 3, \ldots, n$$ equals $$\frac{5(n+1)}{n}$$.

To begin, $$\bar{x} = \frac{n+1}{2}$$.

Next, letting $$n = 2m + 1$$ gives $$\bar{x} = m + 1$$.

The deviations $$|x_i - \bar{x}|$$ are: $$m, m-1, \ldots, 1, 0, 1, \ldots, m$$.

Hence, the sum of deviations is $$2(1 + 2 + \cdots + m) = 2 \cdot \frac{m(m+1)}{2} = m(m+1)$$.

Therefore, the mean deviation is $$\text{MD} = \frac{m(m+1)}{n} = \frac{m(m+1)}{2m+1}$$.

Equating this to $$\frac{5(n+1)}{n} = \frac{5(2m+2)}{2m+1} = \frac{10(m+1)}{2m+1}$$ and noting that $$m + 1 \neq 0$$ allows division by $$\frac{m+1}{2m+1}$$, which yields $$m = 10$$.

Therefore, $$n = 2m + 1 = 21$$.

The answer is $$\boxed{21}$$.