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$N=10a+b$ where $b$ is a non-zero single digit positive integer and $a$ is an $m$ digits positive integer

If we shift $b$ to the first digit's place we get

$2N=10^mb+a$

Combining the two equations

$10^mb+a=20a+2b \iff 19a=10^mb-2b$

$\Longrightarrow a=\dfrac{b(10^m-2)}{19}$

Since $a$ is an integer, $19$ must divide the numerator of the fraction. Since $1\leq b\leq9$ , $19$ must divide $10^m-2$

The smallest $m$ such that $10^m\equiv 2\mod{19}$ is $17$ , plugging in

$a=\dfrac{b(10^{17}-2)}{19}=5263157894736842b$

Since $a$ is a $17$ digits number, $b$ is at least $2$ . $b=2$ gives

$a=5263157894736842\cdot 2 =10526315789473684$

This is a solution since

$N=105263157894736842$

$2N=210526315789473684$

So the smallest $N$ is

$\boxed{105263157894736842}$