Are there numbers $n$ such that the period of $1/p$ is not $n$ for any prime $p$? Based on experimental data, $1$ million primes, I saw that there are no primes $p$ such that the length $l$ of the period of $1/p$ is $19, 23, 24, 36, 38, 39, 47$. Further, for $19$, I have verified that there is no solution in the first $10^9$ primes.
Question 1: Are there numbers $n$ such that the period of $1/p$ is not $n$ for any prime $p$?
Related question: What is the average of ratio of the length of prime periods of $1/p$?
Source code
k = 1
p = 2
step = target = 10^6
found = False

while found == False:
    if p%19 == 1:
        d = divisors(p-1)
        l = len(d)
        i = 1
        while i < l:
            e = d[i]
            if e > 19:
                break
            elif (10^e)%p == 1:
                print(k,p,e)
                break
            elif (10^19)%p == 1:
                found = True
                print('found',k,p,e)
            i = i + 1
    
    if k >= target:
        print(k,p)
        target = target + step
        
    p = next_prime(p)
    k = k + 1

 A: The period of $1/p$ (in base $b \geq 2$ coprime to $p$) is the multiplicative order of $b \in \mathbb{F}_p^{\times}$. Thus if $p$ is a primitive divisor of $b^n-1$ (which, by Zsigmondy’s theorem, always exists except if $(b,n)$ is $(2^t-1,2)$ or $(2,6)$), then $1/p$ has period $n$ in base $b$.
This shows that the answer to your question is “no, every period occurs”.
More generally, the only pairs $(b,n)$ such that there is no prime $p$ coprime to $b$ such that the base-$b$ expansion of $1/p$ has period $n$ are $(2,6)$, $(s,2)$ with $s+1$ a power of two, and $(2,1)$.
A: For $19$, you didn't look far enough.  Try $p=1111111111111111111$.
Why?  (In response to comment.)  Well, the length of the repeating part of $1/p$ is the order of $10$ modulo $p$.  So we search for prime factors of $10^{19}-1$.  Turns out it is just $3^2$ times $p$ above.
If we want to find $p$ such that $1/p$ has repeating part of length $n$, we can look for prime factors of $10^n-1$.  Any such $p$ will have repeating length a factor of $n$, so if $n$ is prime we are guaranteed to succeed.  If $n$ is composite, maybe not.
For $n=24$, calculation shows that $10^n-1$ has $10$ prime factors, and the order of $10$ is $24$ for just one of them: $1/99990001$ has period $24$.  I would expect that to do this in general requires more sophisticated methods, as in @Mindlack's answer.
