Find all natural numbers $n$, such that $\min_{k\in \mathbb{N}}(k^2+[n/k^2])=1991$. ($[n/k^2]$ denotes the integer part of $n/k^2$.)
1. We start by defining the function \( f_k(n) = k^2 + \left\lfloor \frac{n}{k^2} \right\rfloor \). We need to find the natural number \( n \) such that \( \min_{k \in \mathbb{N}} f_k(n) = 1991 \).

2. Consider the function \( g_k(n) = \frac{f_{k+1}(n)}{f_k(n)} \). We analyze the cases where \( g_k(n) \geq 1 \) and \( g_{k-1}(n) \leq 1 \).

3. From \( g_k(n) \geq 1 \), we get:
   \[
   (k+1)^2 + \left\lfloor \frac{n}{(k+1)^2} \right\rfloor \geq k^2 + \left\lfloor \frac{n}{k^2} \right\rfloor
   \]
   This implies:
   \[
   (k+1)^2 \geq k^2 + \left\lfloor \frac{n}{k^2} \right\rfloor - \left\lfloor \frac{n}{(k+1)^2} \right\rfloor
   \]
   Simplifying, we get:
   \[
   k^2(k+1)^2 \geq n
   \]

4. From \( g_{k-1}(n) \leq 1 \), we get:
   \[
   (k-1)^2 + \left\lfloor \frac{n}{(k-1)^2} \right\rfloor \leq k^2 + \left\lfloor \frac{n}{k^2} \right\rfloor
   \]
   This implies:
   \[
   (k-1)^2 \leq k^2 + \left\lfloor \frac{n}{k^2} \right\rfloor - \left\lfloor \frac{n}{(k-1)^2} \right\rfloor
   \]
   Simplifying, we get:
   \[
   n \geq k^2(k-1)^2
   \]

5. Combining these inequalities, we have:
   \[
   k^2(k-1)^2 \leq n \leq k^2(k+1)^2
   \]

6. Let \( i \) be the positive integer satisfying the above inequality. Then:
   \[
   i^2 + \left\lfloor \frac{n}{i^2} \right\rfloor = f_i(n) = \min_{k \in \mathbb{N}} f_k(n) = 1991
   \]

7. From the given condition, we have:
   \[
   \begin{cases}
   i^2(i-1)^2 \leq n \leq i^2(i+1)^2 \\
   (1991 - i^2)i^2 \leq n \leq i^2(1992 - i^2) - 1
   \end{cases}
   \]

8. We analyze the first inequality:
   \[
   i^2(i-1)^2 \leq (1991 - i^2)i^2 \Rightarrow i \leq 32
   \]

9. For \( i \leq 31 \), we need to check:
   \[
   i^2(i+1)^2 \leq i^2(1992 - i^2) - 1
   \]
   This simplifies to:
   \[
   (1991 - i^2)i^2 \leq i^2(i+1)^2
   \]
   This is a contradiction for \( i \leq 31 \).

10. Thus, \( i = 32 \). We check:
    \[
    i^2(i+1)^2 \geq i^2(1992 - i^2) - 1
    \]
    Hence:
    \[
    990208 \leq n \leq 991231
    \]

The final answer is \( \boxed{ n \in \mathbb{N}, 990208 \leq n \leq 991231 } \).