10 The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{1}=1, a_{n+1}=a_{n}+\frac{1}{\left[a_{n}\right]}, n \geqslant 1,
$$

where $\left[a_{n}\right]$ is the integer part of $a_{n}$. For which values of $n$ is the inequality $a_{n}>20$ necessarily true?
10 Write out the first several terms of the sequence $\left\{a_{n}\right\}$:
$$
\begin{array}{c}
a_{1}=1, a_{2}=2, a_{3}=2 \frac{1}{2}, a_{4}=3, a_{5}=3 \frac{1}{3}, a_{6}=3 \frac{2}{3}, \\
a_{7}=4, a_{8}=4 \frac{1}{4}, \cdots .
\end{array}
$$

It can be observed: The sequence is monotonically increasing, and the general term has the form $m+\frac{k}{m}, 0 \leqslant k \leqslant m-1$. We will now prove that this is indeed the case.
When $n=1$, it is clearly true. If $a_{n}=m+\frac{k}{m},(0 \leqslant k \leqslant m-1)$, then we have $\left[a_{n}\right]=m$, thus
$$
a_{n+1}=a_{n}+\frac{1}{\left[a_{n}\right]}=m+\frac{k}{m}+\frac{1}{m}=m+\frac{k+1}{m} .
$$

Therefore, when $k=m-1$, we get $a_{n+1}=m+1$; and when $1 \leqslant k191$, $a_{n}>20$.