Problem 1988

a. Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality
$$q \le \frac{2n + 1}{3} \cdot k.$$
The main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three lines in which it stands, and take into account the cell on which it stands three times, you get $2n + 1$. Let us denote by $S$ this number summed over all checkers. We have
$$S = (2n + 1)q.$$

Now let's find this amount based on the location of the checkers along the lines. Let there be $x_1, \dots, x_n$ checkers in the $n$ horizontal lines, $y_1, \dots, y_n$ checkers in the $n$ lines with angle $60^\circ$, and in the $n$ lines with angle $120^\circ$ there are $z_1, \dots, z_n$ checkers, as shown in Figure 1.

The $i$-th line has exactly $i$ cells, and each of them is in the same line with $x_i, y_i$ or $z_i$ checkers from that line depending on the angle. Therefore,
$$(3) \qquad S = 1 \cdot (x_1 + y_1 + z_1) + 2 \cdot (x_2 + y_2 + z_2) + \dots + n \cdot (x_n + y_n + z_n).$$
Since there are $q$ checkers in total, but there are no more than $k$ checkers in any row, we have the following conditions:
$$x_1 + \dots + x_n = y_1 + \dots + y_n = z_1 + \dots + z_n = q, \\ 0 \le x_i, y_i, z_i \le \min(i, k) \quad \text{for any } 1 \le i \le n.$$
Let $q = mk + r$, where $0 \le r < k$ is the remainder when divided by $k$. It is clear that the expression (3) is maximized when the sums $x_i + y_i + z_i$ with higher coefficients are the maximum possible (that is, equal to $3k$). Hence, under these conditions we get
$$\begin{aligned} (2n + 1)q &= (x_1 + y_1 + z_1) + 2(x_2 + y_2 + z_2) + \dots + n(x_n + y_n + z_n) \\ &\le n \cdot 3k + (n - 1) \cdot 3k + \dots + (n - m + 1) \cdot 3k + (n - m) \cdot 3r \\ &\le \frac{2n - m + 1}{2} m \cdot 3k + (n - m) \cdot 3r \\ &\le \frac{6nmk - 3m^2k + 3mk}{2} + 3nr - 3mr \\ &\le \frac{6n(mk + r) - 3m^2k + 3(mk + r) - 3r - 6mr}{2} \\ &\le \frac{(6n + 3)q - 3m^2k - 3r - 6mr}{2}. \end{aligned}$$
Moving $(2n + 1)q$ to the right side, multiplying the inequality by $2k$ and substituting $mk = q - r$, we arrive at a quadratic inequality for $q$:
$$\begin{aligned} 0 &\le (6n + 3 - 4n - 2)kq - 3(mk)^2 - 3rk - 6mkr \\ &= (2n + 1)kq - 3(q - r)^2 - 6(q - r)r - 3rk \\ &= (2n + 1)kq - 3q^2 + 6qr - 3r^2 - 6qr + 6r^2 - 3rk \\ &= -3q^2 + (2n + 1)kq + 3r(r - k), \end{aligned}$$
from which we obtain
$$3q^2 \le (2n + 1)kq + 3r(r - k) \le (2n + 1)kq, \quad \text{that is} \quad q \le \frac{2n + 1}{3}k,$$
because $r - k < 0$. Since $q$ is an integer, we get $q \le \left\lfloor \frac{2n+1}{3}k \right\rfloor$, as required.

b. We have $T(1, n) = \left\lfloor \frac{2n+1}{3} \right\rfloor$ and $T(2, n) = \left\lfloor \frac{4n+2}{3} \right\rfloor$. Placements of checkers, when equality is achieved in the last inequality, for $k=1$ and $k=2$, are shown in figures 2, 3 and 4.