We are given the expression

$$\left( n_1^{2020} + n_2^{2019} \right)\left( n_2^{2020} + n_3^{2019} \right) \cdots \left( n_{2021}^{2020} + n_{2022}^{2019} \right)\left( n_{2022}^{2020} + n_1^{2019} \right)$$

and need to determine if it can be a power of $11$, i.e., $11^k$ for some $k \in \mathbb{N}$.

To approach this problem, we can use modular arithmetic. Consider the expression modulo $11$:

Each term in the product is of the form $n_{i}^{2020} + n_{i+1}^{2019}$. For each positive integer $n_i$ not divisible by $11$, one of $n_i^{2020}$, $n_{i+1}^{2019}$ will not necessarily be $0 \mod 11$. Note that Fermat's Little Theorem tells us $n_i^{10} \equiv 1 \pmod{11}$ (if $n_i$ is not divisible by $11$), which can extend to show a periodic cycle for the powers involved:

1. $n_i^{2020} = (n_i^{10})^{202} \equiv 1^{202} \equiv 1 \pmod{11}$.
2. $n_i^{2019} = n_i^{10 \times 201} \cdot n_i^{9} \equiv 1^{201} \cdot n_i^{9} \equiv n_i^{9} \pmod{11}$.

Now, observe the expression modulo $11$:
$$n_i^{2020} + n_{i+1}^{2019} \equiv 1 + n_{i+1}^9 \pmod{11}$$

To be a power of $11$, if the entire product could somehow equate to $0 \pmod{11}$, each term would need to individually be divisible by $11$. Otherwise, no complete cancellation or modulus-induced zero can occur. Therefore, each $n_i^{2020} + n_{i+1}^{2019} \equiv 0 \pmod{11}$.

For $n_i^{2020} + n_{i+1}^{2019} \equiv 0 \pmod{11}$, we must have:
$$n_{i+1}^9 \equiv -1 \pmod{11}$$

The power residue $n_{i+1}^9 \equiv -1 \pmod{11}$ introduces inconsistency because:
- $n_i^9 \equiv b \pmod{11}$ implies that there is no integer that squares to $-1$ (since $-1$ is not a quadratic residue mod $11$).

Thus, by attempt to set $n_i^{2020} + n_{i+1}^{2019} = 0 \pmod{11}$ repeatedly contradictions to properties of mod can appear.

Hence, the conclusion is that it is impossible for the given product to be an integer power of $11$ for any choice of positive integers $n_1, n_2, \dots, n_{2022}$.

Therefore, the answer is:
$$\boxed{\text{No}}$$