If a function $$f$$ satisfies $$f(m + n) = f(m) + f(n)$$ for all $$m, n \in \mathbb{N}$$ and $$f(1) = 1$$, then the largest natural number $$\lambda$$ such that $$\sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2$$ is equal to _________

Given: $$f(m+n)=f(m)+f(n)$$ for every $$m,n\in\mathbb{N}$$ and $$f(1)=1$$.

Step 1: Determine $$f(n)$$ for all $$n\in\mathbb{N}$$.
Put $$m=n=1$$ in the functional equation: $$f(2)=f(1)+f(1)=1+1=2$$.
Assume $$f(k)=k$$ for some $$k\in\mathbb{N}$$. Then

$$f(k+1)=f(k)+f(1)=k+1$$

Thus, by mathematical induction, $$f(n)=n$$ for every natural number $$n$$.

Step 2: Express the required sum in terms of $$\lambda$$.
Because $$f(n)=n$$,

$$\sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022} (\lambda+k)$$

Break the sum into two parts:
$$\sum_{k=1}^{2022} \lambda = 2022\lambda$$
$$\sum_{k=1}^{2022} k=\frac{2022\times2023}{2}$$

Therefore,
$$\sum_{k=1}^{2022} f(\lambda+k)=2022\lambda+\frac{2022\times2023}{2}$$ $$-(1)$$

Step 3: Impose the given inequality.
According to the question,

$$2022\lambda+\frac{2022\times2023}{2}\;\le\;(2022)^2$$ $$-(2)$$

Divide every term of $$-(2)$$ by $$2022$$ (note that $$2022\gt0$$):

$$\lambda+\frac{2023}{2}\;\le\;2022$$

Step 4: Solve for $$\lambda$$.
$$\lambda\;\le\;2022-\frac{2023}{2}=2022-1011.5=1010.5$$

The largest natural number not exceeding $$1010.5$$ is $$1010$$.

Answer: The greatest natural number $$\lambda$$ satisfying the given condition is $$1010$$.