For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which $\{(a_1)^1 \times (a_2)^2 \times \ldots \times (a_{20})^{20}\} < \{a_{21} \times a_{22} \times \ldots \times a_{(20+m)}\}$, is

Given $a_k = 3^k$, so $a_1 = 3, a_2 = 9, a_3 = 27, \ldots$

We need the smallest natural number $m$ such that:

$(a_1)^1 \cdot (a_2)^2 \cdot (a_3)^3 \cdots (a_{20})^{20} < a_{21} \cdot a_{22} \cdots a_{20+m}$

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Since $a_k = 3^k$, each term on the LHS becomes:

$(a_k)^k = (3^k)^k = 3^{k^2}$

So LHS $= 3^{1^2} \cdot 3^{2^2} \cdot 3^{3^2} \cdots 3^{20^2}$

$= 3^{1^2 + 2^2 + 3^2 + \cdots + 20^2}$

Using the formula $\displaystyle\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$:

$1^2 + 2^2 + \cdots + 20^2 = \dfrac{20 \times 21 \times 41}{6} = 2870$

So LHS $= 3^{2870}$

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Each term on the RHS is $a_{20+i} = 3^{20+i}$, where $i$ goes from $1$ to $m$.

RHS $= 3^{21} \cdot 3^{22} \cdots 3^{20+m}$

$= 3^{21 + 22 + \cdots + (20+m)}$

$= 3^{\sum_{i=1}^{m}(20+i)}$

$= 3^{20m + \frac{m(m+1)}{2}}$

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The inequality becomes:

$3^{2870} < 3^{20m + \frac{m(m+1)}{2}}$

Since the base $3 > 1$, the exponents follow the same inequality:

$2870 < 20m + \dfrac{m(m+1)}{2}$

$5740 < 40m + m(m+1)$

$5740 < m^2 + 41m$

$m^2 + 41m - 5740 > 0$

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Using the quadratic formula with $m^2 + 41m - 5740 = 0$:

$m = \dfrac{-41 + \sqrt{41^2 + 4 \times 5740}}{2}$

$= \dfrac{-41 + \sqrt{1681 + 22960}}{2}$

$= \dfrac{-41 + \sqrt{24641}}{2}$

$= \dfrac{-41 + 156.99\ldots}{2}$

$\approx \dfrac{116}{2} = 58$

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Checking $m = 57$: $57^2 + 41(57) = 3249 + 2337 = 5586 < 5740$ (does not satisfy)

Checking $m = 58$: $58^2 + 41(58) = 3364 + 2378 = 5742 > 5740$ (satisfies)

The smallest natural number $m$ is $\boxed{58}$.