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$$forall iin {1,2,cdots, k}, n_iinmathbb{N}$$ $$sum_{i=1}^k n_i =n$$
then
$$sum_{i=1}^k n_i^2leq n^2-(k-1)(2n-k)$$

Like comment, If we apply induction,

$i) k=2$

$n_1+n_2=nland n_1,n_2in mathbb{N}$

then
$$n_1^2+n_2^2=n_1^2+(n-n_1)^2=n^2-2n_1n+2n_1^2=2(n_1-frac{n}{2})^2+frac{n^2}{2}$$

Actually this have maximum if $|n_1-frac{n}{2}|$ is maximum(when $n_1=1 lor n_1=n-1$)
therefore $$n_1^2+n_2^2leq 1^2+(n-1)^2= n^2-1*(2n-2)$$

ii)for induction,

suppose $sum_{i=1}^k n_i=nland n_iin mathbb{N}$
imply
$$sum_{i=1}^k n_i^2leq n^2-(k-1)(2n-k)$$

and

claim:$sum_{i=1}^{k+1} n_i=nland n_iin mathbb{N}$
imply
$$sum_{i=1}^{k+1} n_i^2leq n^2-k(2n-k-1)$$
$$sum_{i=1}^{k+1} n_i^2=sum_{i=1}^k n_i^2+n_{k+1}^2leq (n-n_{k+1})^2-(k-1)(2n-2n_{k+1}-k)+n_{k+1}^2$$
$$=2n_{k+1}^2-2n_{k+1}(n-k+1)+n^2-2n(k-1)+k(k-1)$$
$$=2(n_{k+1}-frac{n-k+1}{2})^2-frac{(n-k+1)^2}{2}+n^2-2n(k-1)+k(k-1)$$

Actually It have maximum if $|n_{k+1}-frac{n-k+1}{2}|$ is maximum, (at $n_{k+1}=1lor n-k$) therefore
$$leq frac{(n-k-1)^2}{2}-frac{(n-k+1)}{2}+n^2-2n(k-1)+k(k-1)$$
$$=-2(n-k)+n^2-2n(k-1)+(k-1)k$$
$$=n^2-k(2n-k-1)$$

therefore
$$sum_{i=1}^k n_i^2leq n^2-(k-1)(2n-k)$$ and equality occur when $forall i= 1,2,cdots,k-1 ,n_i=1,n_k=n-k+1$

Let $m = min n_i$ and $M = max n_i$. We are told $n_i ge 1$ (i.e. $mathbb{N}$ stands for the set of positive integers), so $m ge 1$ and hence
$$M le n - (k-1)m le n - k + 1$$

This leads to

$$begin{align}
sum_i n_i^2 &= sum_i ((n_i+1)(n_i-1) + 1)
le (M+1)sum_i (n_i - 1) + k\
&le (n - k + 2)(n-k) + k
= n^2 - (k-1)(2n-k)
end{align}tag{*1}
$$
This is the inequality we want to prove. We can improve this inequality by
using the actual maximum $M$ and minimum $m$. We have

$$begin{align}
sum_i n_i^2 &= sum_i ((n_i+m)(n_i-m) + m^2)
le (M+m)sum_i (n_i - m) + km^2\
&le (M+m)(n - km) + km^2
= (M+m)n - kMm
end{align}
$$
Let $mu = frac{n}{k}$ and $sigma$ be the mean and standard derivation of $n_i$. Above inequality is equivalent to

$$k (sigma^2 + mu^2) le (M+m)mu k - kMm
quadiffquad sigma^2 le (M - mu)(mu - m)tag{*2}$$

When we replace $M$ and $m$ by other upper/lower bounds for $n_i$, $(M - mu)(mu - m)$ will only getting bigger and inequality $(*2)$ remains valid.
The inequality $(*1)$ is really a special case of this when we replace $M, m$ by $n-k+1$ and $1$.

The inequality on RHS$(*2)$ is known as Bhatia–Davis inequality. Look at its wiki entry for similar bounds.