Suppose we have some set $S = \{1,...,N\}$ and partition this set into subsets $S_i \in S$ such that $S_i \cap S_j = \emptyset$ if $i\neq j$. How many different partitions can make such that the biggest subsize has atleast cardinalty $|S| = k$.
Example
Let $S = \{1,2,3,4\}$ then there us ${4 \choose 4} = 1$ way to partition when $k=4$. There are ${4\choose 3} =4$ ways to partition when $k=3$. However there are ${4 \choose 2} + 3 = 9$ ways to partition when $k = 2$. The extra three comes from the partitions $(\{1,2\}, \{3,4\})$, $(\{1,3\}, \{2,4\})$, $(\{1,4\}, \{2,3\})$.
Is there a general way to calculate the number of partitions?
I am a little confused as to what you are asking. You say you want to count the partitions where the largest part has size at least $k$. However, in the case where $N=4$ and $k=3$, you are only counting the four partitions that look like $(\{a,b,c\},\{d\})$. I think there should be five partitions in this case, since $(\{1,2,3,4\})$ also has largest part whose size is at least $3$.
Let
$g(n,k)=$ the number of partitions of $\{1,\dots,n\}$ whose largest part has at least $k$ elements
$e(n,k)=$ the number of partitions of $\{1,\dots,n\}$ whose largest part has exactly $k$ elements
$\ell(n,k)=$ the number of partitions of $\{1,\dots,n\}$ whose largest part has at most $k$ elements
Basically, you start out asking for $g(n,k)$, but your examples are all computing $e(n,k)$.
I will show you how to compute all three of these functions. Let us start with $\ell(n,k)$. Using the theory of exponential generating functions, you can show $$ \ell(n,k)=[x^n]\exp\left(\sum_{i=1}^{k}\frac{x^i}{i!}\right) $$ The notation $[x^n]f(x)$ means "the coefficient of $x^n$ in the power series $f(x)$." This can be computed using any computer algebra system, like Mathematica or Maple. We can also give a recursive equation for $\ell(n,k)$, which allows you to compute $\ell(n,k)$ using dynamic programming and the help of a computer. $$ \ell(n+1,k)=\sum_{i=0}^{k-1} \binom {n}i \ell(n-i,k) $$ Now, we can leverage this to compute $g(n,k)$. We have that $$ g(n,k)+\ell(n,k-1)=\text{total number of partitions} =B_n, $$ where $B_n$ is the $n^\text{th}$ Bell number. You can compute the Bell numbers using the exponential generating function equation $$B_n=[x^n]\exp(\exp (x)-1),$$ or the recursive equation $$ B_{n+1}=\sum_{i=0}^n\binom ni B_i. $$ Finally, we use either $\ell(n,k)$ or $g(n,k)$ to compute $e(n,k)$: $$ e(n,k)=\ell(n,k)-\ell(n,k-1)=g(n,k)-g(n,k+1). $$