Find all pairs of natural numbers $(\alpha,\beta)$ for which, if $\delta$ is the greatest common divisor of $\alpha,\beta$, and $\varDelta$ is the least common multiple of $\alpha,\beta$, then 
\[ \delta + \Delta = 4(\alpha + \beta) + 2021\]

1. **Define Variables and Rewrite the Equation:**
   Let $\alpha = d \cdot x$ and $\beta = d \cdot y$, where $d = \gcd(\alpha, \beta)$ and $x, y$ are coprime integers. The least common multiple $\Delta$ of $\alpha$ and $\beta$ is given by:
   \[
   \Delta = \frac{\alpha \beta}{\gcd(\alpha, \beta)} = \frac{d \cdot x \cdot d \cdot y}{d} = d \cdot x \cdot y
   \]
   The given equation is:
   \[
   \delta + \Delta = 4(\alpha + \beta) + 2021
   \]
   Substituting $\delta = d$ and $\Delta = dxy$, we get:
   \[
   d + dxy = 4d(x + y) + 2021
   \]
   Simplifying, we have:
   \[
   d(1 + xy) = 4d(x + y) + 2021
   \]
   Dividing both sides by $d$ (assuming $d \neq 0$), we get:
   \[
   1 + xy = 4(x + y) + \frac{2021}{d}
   \]
   Rearranging, we obtain:
   \[
   d(xy - 4x - 4y + 15) = 2021
   \]
   This can be rewritten as:
   \[
   d(x-4)(y-4) = 2021 + 15d
   \]

2. **Determine Possible Values of $d$:**
   Since $d$ must be a divisor of 2021, we find the divisors of 2021. The prime factorization of 2021 is:
   \[
   2021 = 43 \times 47
   \]
   Therefore, the divisors of 2021 are:
   \[
   d \in \{1, 43, 47, 2021\}
   \]

3. **Solve for Each Possible Value of $d$:**
   - **For $d = 1$:**
     \[
     (x-4)(y-4) = 2036
     \]
     We need to find pairs $(x, y)$ such that $x$ and $y$ are coprime and $(x-4)(y-4) = 2036$. The factor pairs of 2036 are:
     \[
     (1, 2036), (2, 1018), (4, 509), (7, 292), (14, 146), (28, 73)
     \]
     Checking for coprime pairs:
     \[
     (x, y) = (8, 513), (513, 8)
     \]

   - **For $d = 43:**
     \[
     (x-4)(y-4) = 62
     \]
     The factor pairs of 62 are:
     \[
     (1, 62), (2, 31)
     \]
     Checking for coprime pairs:
     \[
     (x, y) = (5, 66), (66, 5), (6, 35), (35, 6)
     \]

   - **For $d = 47:**
     \[
     (x-4)(y-4) = 58
     \]
     The factor pairs of 58 are:
     \[
     (1, 58), (2, 29)
     \]
     Checking for coprime pairs:
     \[
     (x, y) = (5, 63), (63, 5)
     \]

   - **For $d = 2021:**
     \[
     (x-4)(y-4) = 16
     \]
     The factor pairs of 16 are:
     \[
     (1, 16), (2, 8), (4, 4)
     \]
     None of these pairs are coprime.

4. **List All Pairs $(\alpha, \beta)$:**
   - For $d = 1$:
     \[
     (\alpha, \beta) = (8, 513), (513, 8)
     \]
   - For $d = 43$:
     \[
     (\alpha, \beta) = (5 \cdot 43, 66 \cdot 43) = (215, 2838), (2838, 215), (6 \cdot 43, 35 \cdot 43) = (258, 1505), (1505, 258)
     \]
   - For $d = 47$:
     \[
     (\alpha, \beta) = (5 \cdot 47, 63 \cdot 47) = (235, 2961), (2961, 235)
     \]

The final answer is $\boxed{(8, 513), (513, 8), (215, 2838), (2838, 215), (258, 1505), (1505, 258), (235, 2961), (2961, 235)}$