15. Xiaohua wrote a computer program. One minute after the program runs, some bubbles first appear on the computer screen, and at each subsequent integer minute, the same number of new bubbles appear. Half a minute after the 11th appearance of bubbles, one bubble bursts. Thereafter, every minute, more bubbles burst, with the number increasing by 1 each minute (i.e., half a minute after the 12th appearance of bubbles, 2 bubbles burst...). At a certain moment, the total number of burst bubbles is exactly equal to the total number of bubbles that have appeared on the screen, meaning all bubbles have disappeared at this moment. How many bubbles were there at most on the screen at any given time during the entire process of the program running?
【Solution】Solution: Let the number of bubbles appearing each time be $k$, and after the $m$-th bubble burst, the total number of burst bubbles is exactly equal to the total number of bubbles that have appeared on the computer screen. Before this, there have already been $m+10$ times of bubble appearances. According to the total number of burst bubbles being exactly equal to the total number of bubbles that have appeared on the computer screen, we get:
$$
(10+m) k=\frac{m(m+1)}{2}
$$

Simplifying, we get: $2(10+m) k=m(m+1)$, since $m, k$ are both positive integers, so
$$
m+10 \mid m^{2}+m-(m+10)(m-9)
$$

We get:
$$
m+10 \mid 90
$$

Therefore, $m+10=15,18,30,45,90$
So, $m=5,8,20,35,80$, and the corresponding $k=1,2,7,14,36$
When the number of burst bubbles is less than the number of new bubbles appearing, the total number of bubbles on the computer screen increases; when the number of burst bubbles is greater than the number of new bubbles appearing, the total number of bubbles on the computer screen decreases; only when the number of burst bubbles equals the number of new bubbles appearing, the total number of bubbles on the computer screen is maximized. Thus, the number of bubbles on the screen is maximized in the first half-minute before the $k$ bubbles burst, at which point the number of bubbles displayed on the computer is (10+k) $k-\frac{k(k-1)}{2}=\frac{k(k+21)}{2}$. The larger the $k$, the more bubbles are displayed. When $k=36$, it is easy to verify that the condition is satisfied, at which point the number of bubbles displayed is 1026.
Therefore, the answer is: 1026.