1. We are given

   \quad \text{for all } m,n\in \mathbb N,$$ and $$f(1)=1.$$

2. Since the relation holds for all natural numbers, we can determine $f(n)$ for every $n\in\mathbb N$.

   Using induction / repeated addition: $f(2)=f(1+1)=f(1)+f(1)=2,$ $f(3)=f(2+1)=f(2)+f(1)=3,$ and in general, $f(n)=n f(1)=n.$ Since $f(1)=1$, we get $f(n)=n \quad \text{for all } n\in\mathbb N.$

3. Now evaluate $\sum_{k=1}^{2022} f(\lambda+k).$ Since $f(n)=n$, $\sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022} (\lambda+k).$

4. Split the sum: $\sum_{k=1}^{2022} (\lambda+k)=\sum_{k=1}^{2022} \lambda+\sum_{k=1}^{2022} k.$ Hence, $=2022\lambda+\frac{2022\cdot 2023}{2}.$

5. The given inequality is $2022\lambda+\frac{2022\cdot 2023}{2}\le (2022)^2.$

6. Simplify: $\frac{2022\cdot 2023}{2}=1011\cdot 2023=2045253,$ and $(2022)^2=4088484.$ So, $2022\lambda+2045253\le 4088484.$ Therefore, $2022\lambda\le 4088484-2045253=2043231.$

7. Divide by $2022$: $\lambda\le \frac{2043231}{2022}.$ Now, $2022\cdot 1010=2042220,$ $2022\cdot 1011=2044242.$ Since $2042220\le 2043231<2044242,$ the largest natural number satisfying the inequality is $\lambda=1010.$

8. Verification: For $\lambda=1010$, $\sum_{k=1}^{2022} (1010+k)=2022\cdot 1010+\frac{2022\cdot 2023}{2}=4087473\le 4088484.$ For $\lambda=1011$, the sum increases by $2022$, giving $4087473+2022=4089495>4088484,$ so $1011$ does not work.

Therefore, the required largest natural number is $\boxed{1010}.$