Given the roots \( z_1, z_2, \ldots, z_{20} \) of the polynomial equation
\[
z^{20} - 4z^{19} + 9z^{18} - 16z^{17} + \dots + 441 = 0,
\]
we need to determine the value of \( \cot \left( \sum_{k = 1}^{20} \operatorname{arccot} z_k \right) \).
First, recall the properties of the \(\arccot\) function and how it relates to complex roots and trigonometric identities. Specifically, for any complex number \( z \),
\[
\operatorname{arccot} z = \arctan \frac{1}{z}.
\]
Next, using the identity for the sum of \(\arctan\) values, we have:
\[
\arctan z_1 + \arctan z_2 + \cdots + \arctan z_{20} = \arctan \left( \frac{z_1 + z_2 + \cdots + z_{20}}{1 - z_1 z_2 \cdots z_{20}} \right).
\]
Given that \(\operatorname{arccot} z_k = \arctan \frac{1}{z_k}\), it follows that:
\[
\sum_{k=1}^{20} \operatorname{arccot} z_k = \sum_{k=1}^{20} \arctan \frac{1}{z_k}.
\]
We use the identity for the sum of arctangents. For simplicity, let's assume that the complex numbers \( z_1, z_2, \ldots, z_{20} \) are such that we can use:
\[
\arctan \frac{1}{z_1} + \arctan \frac{1}{z_2} + \cdots + \arctan \frac{1}{z_{20}} = \arctan \left( \frac{\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_{20}}}{1 - \left( \frac{1}{z_1} \cdot \frac{1}{z_2} \cdots \frac{1}{z_{20}} \right)} \right).
\]
Now, we need to analyze the roots of the given polynomial \( z^{20} - 4z^{19} + 9z^{18} - \cdots + 441 = 0 \).
By Vieta's formulas, the sum of the roots (considering the coefficients of the polynomial) is equal to the coefficient of \( z^{19} \) divided by the leading coefficient (for \( z^{20} \)), which is \( -\frac{-4}{1} = 4 \).
However, for the product of the roots, by Vieta's formulas, the constant term (considering all terms up to the last one) will give us:
\[
z_1 z_2 \cdots z_{20} = (-1)^{20} \times \text{constant term}/\text{leading coefficient} = 441.
\]
Thus, for the cotangent sum, we need the real parts of the roots in terms of trigonometric identities:
\[
\cot \left( \sum_{k=1}^{20} \operatorname{arccot} z_k \right) = \cot \left( \arctan \left( \frac{\sum_{k=1}^{20} \frac{1}{z_k}}{1 - \frac{1}{z_1} \frac{1}{z_2} \cdots \frac{1}{z_{20}}} \right) \right).
\]
Since \(\sum_{k=1}^{20} z_k = 4\) and \(\prod_{k=1}^{20} z_k = 441\),
\[
\sum_{k=1}^{20} \frac{1}{z_k} = \frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_{20}} = \frac{20}{441}.
\]
Then:
\[
1 - \prod_{k=1}^{20} \frac{1}{z_k} = 1 - \frac{1}{441} = \frac{440}{441}.
\]
Substituting in,
\[
\frac{\frac{20}{441}}{\frac{440}{441}} = \frac{20}{440} = \frac{1}{22}.
\]
Therefore, we have
\[
\arctan \left( \frac{1}{22} \right),
\]
which means
\[
\cot \left( \arctan \frac{1}{22} \right) = 22.
\]
Hence, the final answer is:
\[
\boxed{22}.
\]
