Riquinho distributed $R \$ 1000.00$ 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. How much did Bernardo receive?
The money was divided into installments in the form

$$
1+2+3+\cdots+n \leq 1000
$$

Since $1+2+3+\cdots+n$ is the sum $S_{n}$ of the first $n$ natural numbers starting from $a_{1}=1$, we have:

$$
S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}=\frac{(1+n) n}{2} \leq 1000 \Longrightarrow n^{2}+n-2000 \leq 0
$$

We have that

$$
n^{2}+n-2000<0 \quad \text { for values of } n \text { between the roots }
$$

Since the positive solution of $n^{2}+n-2000=0$ is

$$
n=\frac{-1+\sqrt{1+8000}}{2} \simeq 44.22
$$

then $n \leq 44$. Thus, Bernardo received

$$
2+5+8+11+\cdots+44=\frac{(44+2) \cdot 15}{2}=23 \cdot 15=345
$$