Riquinho distributed 1000.00 reais among his friends Antônio, Bernardo, and Carlos in the following manner: he gave, successively, 1 real to Antônio, 2 reais to Bernardo, 3 reais to Carlos, 4 reais to Antônio, 5 reais to Bernardo, etc. What was the amount received by Bernardo?
The 1000 reais of Riquinho were divided into increasing installments starting from 1, such that $1+2+3+\cdots+n \leq 1000$. Since $1+2+3+\cdots+n$ is the sum of the first $n$ integers starting from 1, we have $1+2+3+\cdots+n=\frac{1}{2}(1+n) n$. Thus, we want to find the largest $n$ such that $\frac{1}{2}(1+n) n=1+2+3+\cdots+n \leq 1000$, or in other words, such that $n^{2}+n-2000 \leq 0$.

Since $n^{2}+n-2000$ is equal to -2000 for $n=0$ and the leading coefficient of this quadratic polynomial is $1>0$, we know that the values of $n^{2}+n-2000$ are negative for all $n$ between 0 and the positive root of the quadratic polynomial $x^{2}+x-2000$. By the quadratic formula, the positive root is given by

$$
x=\frac{-1+\sqrt{1+8000}}{2} \approx 44.22
$$

thus $n^{2}+n>2000$ for $n \geq 45$. Therefore, Riquinho distributed only 44 installments. Since Bernardo received the second installment, the fifth installment $(5=2+3)$, the eighth installment $(8=2+2 \times 3)$, and so on, he also received the last one, since $44=2+14 \times 3$, for a total of

$$
2+5+8+11+\cdots+44=\frac{1}{2}(44+2) \times 15=23 \times 15=345 \text { reais. }
$$

Note: After distributing the 44 installments, there are still

$$
1000-\frac{1}{2}(44 \times 45)=1000-990=10
$$

of the 1000 reais of Riquinho.