1. Variance of the first $n$ natural numbers

The first $n$ natural numbers are: $1,2,3,\dots,n$

For the numbers $1$ to $n$:

• Mean is $\bar{x}=\frac{1+2+\cdots+n}{n}=\frac{n+1}{2}$

• Also, $\frac{1^2+2^2+\cdots+n^2}{n}=\frac{n(n+1)(2n+1)}{6n}=\frac{(n+1)(2n+1)}{6}$

Variance is $\sigma^2=\frac{1^2+2^2+\cdots+n^2}{n}-\bar{x}^2$

So, $\sigma^2=\frac{(n+1)(2n+1)}{6}-\left(\frac{n+1}{2}\right)^2$

Simplifying, $\sigma^2=\frac{n^2-1}{12}$

Given variance is $10$: $\frac{n^2-1}{12}=10$ $n^2-1=120$ $n^2=121$ $n=11$

Since $n$ is a natural number, $n=11$.

2. Variance of the first $m$ even natural numbers

The first $m$ even natural numbers are: $2,4,6,\dots,2m$

These are obtained by multiplying $1,2,3,\dots,m$ by $2$.

If a data set is multiplied by a constant $k$, then its variance is multiplied by $k^2$.

Variance of $1,2,3,\dots,m$ is: $\frac{m^2-1}{12}$

Therefore variance of $2,4,6,\dots,2m$ is: $4\cdot \frac{m^2-1}{12}=\frac{m^2-1}{3}$

Given this variance is $16$: $\frac{m^2-1}{3}=16$ $m^2-1=48$ $m^2=49$ $m=7$

Since $m$ is a natural number, $m=7$.

3. Find $m+n$

$m+n=7+11=18$

So the required integer is: $\boxed{18}$