From the sequence of positive integers ranging from 1 to 1993, we removed three consecutive numbers. Taking the arithmetic mean of the remaining terms, we found that it is also an integer. Which are the removed numbers?
Mint is well-known, the sum of integers up to $n$ can be calculated using the formula $S_{n}=\frac{n(n+1)}{2}$. Let's denote the middle number among the three removed numbers as $x$, then

$$
\text { (1) } \frac{\frac{1993.1994}{2}-3 x}{1990}=N \text {, }
$$

where $N$ is an integer. Let's find the values of $x$ for which the equality holds.

Transform (1) into

$$
\frac{\frac{1987021}{2}-3 x}{1990}=N
$$

Notice that the value of $x$ can be at least 2 and at most 1992.

When $x=2$, $N=998.5$.

When $x=1992$, $N=995.9$.

Since $N$ is an integer, and its value decreases as $x$ increases, $N$ can only be 996, 997, or 998. It is easy to verify that only when $N=997$, the value of $x$ is an integer, which is 997.

Therefore, the removed numbers are: $996,997,998$.