1. Given data

We have the numbers: $1,2,3,\dots,n$ with $n$ odd.

The mean deviation about the mean is given as: $\frac{5(n+1)}{n}$

We need to find $n$.

2. Mean of the numbers $1$ to $n$

The mean of the first $n$ natural numbers is: $\bar{x}=\frac{1+2+\cdots+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$

Since $n$ is odd, let: $n=2m+1$ Then: $\bar{x}=\frac{n+1}{2}=m+1$ which is one of the terms.

3. Mean deviation about the mean

Mean deviation about the mean is: $\text{MD}=\frac{1}{n}\sum_{k=1}^n \left|k-\frac{n+1}{2}\right|$

Because the terms are symmetric about the mean, the deviations are: $m,m-1,\dots,1,0,1,\dots,m-1,m$

So, $\sum_{k=1}^n \left|k-\frac{n+1}{2}\right|=2(1+2+\cdots+m)$

Now, $1+2+\cdots+m=\frac{m(m+1)}{2}$ Therefore, $\sum |x_i-\bar{x}|=2\cdot \frac{m(m+1)}{2}=m(m+1)$

Hence, $\text{MD}=\frac{m(m+1)}{n}$

Since $m=\frac{n-1}{2}$, we get: $m+1=\frac{n+1}{2}$ So, $m(m+1)=\frac{n-1}{2}\cdot \frac{n+1}{2}=\frac{n^2-1}{4}$

Thus, $\text{MD}=\frac{n^2-1}{4n}$

4. Use the given condition

Given: $\frac{n^2-1}{4n}=\frac{5(n+1)}{n}$

Multiply both sides by $4n$: $n^2-1=20(n+1)$

$n^2-1=20n+20$

$n^2-20n-21=0$

Factorizing: $n^2-21n+n-21=0$ $n(n-21)+1(n-21)=0$ $(n-21)(n+1)=0$

So, $n=21 \quad \text{or} \quad n=-1$

Since $n$ is a positive integer, $n=21$

5. Final answer

$\boxed{21}$

6. Comparison with stored correct answer

Stored correct answer = $21$.

Our derived answer also is $21$, so they agree.