Square and Cube

$2 \times N=m^2 \\ 3 \times N=n^3$

Multiplying them $6 \times N^2=m^2 n^3=(mn)^2 \times n$

Clearly, $n=6$ So, $3N=216 \implies n=72$ also $2N=144$

So, $N=72$

Since $2N$ is a square and $3N$ is a cube, $N$ has to be of the form $\large 2^a \times 3^b$ , and cannot have another prime factor because then $N$ will not be the smallest.

$\large \Longrightarrow N = 2^a \times 3^b$

$\large \Longrightarrow 2N = 2^{a + 1} \times 3^b \quad \cdots (1)$

$\large \Longrightarrow 3N = 2^a \times 3^{b + 1} \quad \cdots (2)$

Now, $2N$ is a perfect square, it means the both the powers are even $\implies b = 0,2,4,6 \cdots$

and $3N$ is a perfect cube, it means the both the powers are multiple of 3 $\implies a = 0,3,6,9 \cdots$

Here $b \ne 0$ because it will not satisfy equation (2), and $a \ne 0$ because it will not satisfy equation (1)

hence, the smallest values of $a$ and $b$ which satisfy both the equations are $a = 3$ and $b = 2$ .

$\implies N = 2^3 \times 3^2 = 72$

sum of digits = $\boxed {9}$

For $2N$ to be a square, $N$ has to have the format $2\times x^2$ .

For $3N$ to be a cube, $N$ has to have the format $3^2\times y^3$ .

So $N$ definitely has to be a multiple of $2\times3^2=18$

$18$ does have the first format, $2\times x^2$ with $x=3$ , but it does not have the second, $3^2\times y^3$ . It does have the $3^2$ part, but that is multiplied by $2$ , which is not a cube. To keep the $2$ and make it a cube, and the smallest cube possible, it needs to be multiplied by $4$ , which is a square.

$18\times 4=72$ satisfies both conditions and it is the lowest number to do so.

The sum of its digits is $\boxed{9}$ .

Let $2N = p^2$ and $3N = q^3.$ The least value of N that satisfies both conditions is $N = 2^{3}3^{2} = 72$ , which has a digit sum of $\boxed{9}.$