Consider the general term in the expansion of $\left(x^2+\frac{1}{x^3}\right)^n.$ If we choose $r$ times the second term $\dfrac{1}{x^3}$ and $(n-r)$ times the first term $x^2$, then the general term is $T_{r+1}=\binom{n}{r}(x^2)^{n-r}\left(x^{-3}\right)^r.$

Simplify the power of $x$: $T_{r+1}=\binom{n}{r}x^{2(n-r)}x^{-3r}=\binom{n}{r}x^{2n-2r-3r}=\binom{n}{r}x^{2n-5r}.$

We need the coefficient of $x^1$, so the exponent must be $1$: $2n-5r=1.$ Thus, $r=\frac{2n-1}{5}.$ Since $r$ must be an integer, $2n-1$ must be divisible by $5$.

The coefficient of $x$ is then $\binom{n}{r}.$ Given that this coefficient is $\binom{n}{23}$, we need $r=23$ (or possibly $r=n-23$, since $\binom{n}{r}=\binom{n}{n-r}$).

First check $r=23$: $2n-5(23)=1$ $2n-115=1$ $2n=116$ $n=58.$ This is valid.

Now check the symmetric possibility $r=n-23$: From $2n-5r=1,$ substitute $r=n-23$: $2n-5(n-23)=1$ $2n-5n+115=1$ $-3n=-114$ $n=38.$ This is also valid, because then $r=n-23=15,$ and $\binom{n}{r}=\binom{38}{15}=\binom{38}{23}.$

We need the smallest natural number $n$. Between $58$ and $38$, the smaller is $n=38.$

Therefore, the correct option is $\boxed{38}.$