Let's determine the smallest value of $\left|36^{m}-5^{n}\right|$, where $m$ and $n$ are positive integers.
For every positive integer $m$ and $n$, the last digit of $36^{m}$ in the decimal system is 6, and the last digit of $5^{n}$ is 5. Thus, the last digit of $\left|36^{m}-5^{n}\right|$ is 1 when $36^{m}>5^{n}$, and 9 when $36^{m}<5^{n}$.

The value of $\left|36^{m}-5^{n}\right|$ cannot be 1, because in that case $36^{m}-5^{n}=1$, which means $36^{m}-1=5^{n}$. However, $36^{m}-1$ is divisible by $36-1=35$, and thus divisible by 7, whereas $5^{n}$ is not. Similarly, it cannot be 9, since from $36^{m}-5^{n}=-9$ we get $36^{m}+9=5^{n}$, and the left side is divisible by 9, while $5^{n}$ is not.

The next possible value is 11, and this expression does indeed take on this value for $m=1$ and $n=2$. Therefore, the solution to the problem is: 11.