[THOUGHT] To compute the fourth moment $\mathbb{E}[N^4]$ for $N \sim \mathrm{Binomial}(20, 1/2)$, we use the formula involving Stirling numbers of the second kind. The fourth moment is given by:

where $S(4, k)$ are the Stirling numbers of the second kind, $n^{\underline{k}}$ is the falling factorial, and $p = 1/2$. The Stirling numbers for $k=1$ to $4$ are $1, 7, 6, 1$ respectively. Substituting $n=20$ and $p=1/2$, we calculate each term and sum them up. Multiplying the result by 1000 gives $K$.

[/THOUGHT] [PYTHON]

Calculate the fourth moment using Stirling numbers of the second kind

n = 20 p = 1/2

Stirling numbers of the second kind for k=1 to 4

S = [1, 7, 6, 1]

Falling factorials

falling = [ n, # n n * (n - 1), # n(n-1) n * (n - 1) * (n - 2), # n(n-1)(n-2) n * (n - 1) * (n - 2) * (n - 3) # n(n-1)(n-2)(n-3) ]

Calculate each term and sum

E_N4 = sum(S[k] * falling[k] * (p ** (k + 1)) for k in range(4)) K = int(E_N4 * 1000) print(K) [/PYTHON]

[THOUGHT] The Python code computed $K = 1000 \cdot \mathbb{E}[N^4] = 13072500$ using the Stirling numbers approach. The output is valid and matches the required integer format. [/THOUGHT] $\boxed{13072500}$