9・86 In a $19 \times 89$ rectangular grid, what is the maximum number of small squares that can be colored black so that in any $2 \times 2$ square, there are at most two black squares?
［Solution］If we color all the odd-numbered rows of the 19 rows of squares in the rectangle as shown in the right figure, it obviously meets the requirements of the problem and there are 890 black squares in total. Therefore, the maximum number of black squares required by the problem is no less than 890.

Next, we will prove that under the conditions of the problem, the number of black squares in the table cannot exceed 890. For this purpose, we first point out that it is easy to prove by mathematical induction that in a $2 \times (2n+1)$ rectangular grid, under the conditions of the problem, the table can have at most $2(n+1)$ black squares, and only by coloring in the way shown in Figure (b) can the table have exactly $2(n+1)$ black squares.

We then use mathematical induction to prove that in a $(2m+1) \times (2n+1)$ rectangular grid $(n \geqslant m)$, if some squares are colored black and meet the conditions of the problem, then the number of black squares is no more than $(m+1)(2n+1)$.
When $m=0$, the proposition is obviously true. Suppose the proposition holds for $m=k$. When $m=k+1$, in a $(2k+3) \times (2n+1)$ rectangular grid, we separate a $2 \times (2n+1)$ strip at the top. If the number of black squares in the strip does not exceed $2(n+1)-1=2n+1$, then by the inductive hypothesis, the number of black squares in the $(2k+1) \times (2n+1)$ grid below does not exceed $(k+1)(2n+1)$, so the total number of black squares in the entire $(2k+3) \times (2n+1)$ grid does not exceed $(k+2)(2n+1)$, i.e., the proposition holds. If the number of black squares in the strip is no less than $2(n+1)$, then by the discussion above, the number of black squares is exactly $2(n+1)$ and can only be colored as shown in Figure (b). At this time, we can cut off the first row of the $(2k+3) \times (2n+1)$ rectangular grid, which has exactly $(n+1)$ black squares. The remaining $(2k+2) \times (2n+1)$ rectangular grid is divided into $k+1$ strips of $2 \times (2n+1)$, so each strip has at most $2(n+1)$ black squares. Therefore, the total number of black squares in the entire grid does not exceed $(n+1) + (2k+2)(n+1) = (2k+3)(n+1) \leqslant (k+2)(2n+1)$, i.e., the proposition holds for $m=k+1$.
In summary, in the coloring method that meets the conditions of the problem, the table can have at most 890 black squares.