jmc

If $n$ is a two-digit number, then we can write $n$ in the form $10a+b$, where $a$ and $b$ are digits. Then the last digit of $n^2$ is the same as the last digit of $b^2$.

The last digit of $n^2$ is 1. We know that $b$ is a digit from 0 to 9. Checking these digits, we find that the units digit of $b^2$ is 1 only for $b=1$ and $b=9$.

If $b=1$, then $n=10a+1$, so $$n^2=100a^2+20a+1.$$ The last two digits of $100a^2$ are 00, so we want the last two digits of $20a$ to be 00. This occurs only for the digits $a=0$ and $a=5$, but we reject $a=0$ because we want a two-digit number. This leads to the solution $n=51$.

If $b=9$, then $n=10a+9$, so $$n^2=100a^2+180a+81=100a^2+100a+80a+81.$$ The last two digits of $100a^2+100a$ are 00, so we want the last two digits of $80a+81$ to be 01. In other words, we want the last digit of $8a+8$ to be 0. This only occurs for the digits $a=4$ and $a=9$. This leads to the solutions $n=49$ and $n=99$.

Therefore, the sum of all two-digit positive integers whose squares end with the digits 01 is $51+49+99=\boxed{199}$.