To solve this problem, we need to find the smallest positive integer $n$ such that all 73 fractions of the form $\frac{k}{n+k+20}$ for $k = 19, 20, \ldots, 91$ are irreducible. A fraction $\frac{a}{b}$ is irreducible if and only if $\gcd(a, b) = 1$.

For the fractions $\frac{k}{n+k+20}$ to be irreducible, we need:

$$\gcd(k, n+k+20) = 1$$

Simplifying the statement $\gcd(k, n+k+20)$, we observe that:

$$\gcd(k, n+k+20) = \gcd(k, n+20)$$

Using the Euclidean Algorithm, we understand this stems from the fact that if $\gcd(a, b) = \gcd(a, b - qa)$ for any integer $q$.

Given the fraction range, each $k$ increments by 1 starting from 19. Therefore, we require:

$$\gcd(k, n+20) = 1 \quad \text{for} \quad k = 19, 20, \ldots, 91$$

Notice that this results in 73 simultaneous conditions, corresponding to each value of $k$.

The common task here is discerning $n$ such that $n+20$ is coprime to each $k \in \{19, 20, \ldots, 91\}$.

We observe that for each $k$, the main requirement is that $n + 20$ needs to avoid any prime factors present within the sequence $\{19, 20, \ldots, 91\}$.

The numbers $k = 19, 20, 21, \ldots, 91$ encompass several prime numbers such as $19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89$. Each $n + 20$ must ensure it is not divisible by any of these prime numbers. The simplest way is to let $n + 20$ exceed the largest of these numbers, for clear coprime status.

The largest prime in this list is 89. Thus, $n + 20$ should be at least no less than 89, giving room for searching the smallest $n$.

Therefore, the minimum feasible $n + 20$ which satisfies all coprime conditions can be tested progressively upwards from 90, knowing large enough clear routes.

Finally, solving for the smallest feasible $n$:

$n + 20 = 115 \quad \Rightarrow \quad n = 95$.

Thus, the smallest positive integer $n$ that meets the condition of being coprime across the entire defined range is:

$$\boxed{95}$$