First, note that we want $n$ to be very close, but more than, $\frac{m}{3}.$ We take cases on the residue of $m$ modulo $3$: (i) If $m \equiv 0 \pmod{3},$ then by the reasoning above, the smallest positive value of $k$ is $3 - \frac{m}{\frac{m}{3} + 1} = 3 - 3 \cdot \frac{m}{m+3}.$ Clearly, $\frac{m}{m+3}$ gets bigger and bigger, approaching $1$, as $m$ increases, so to minimize $3 - 3 \cdot \frac{m}{m+3},$ we want $m$ to be the largest multiple of $3$ less than or equal to $2000$ which is $1998.$ If $m=1998,$ this quantity is $\frac{3}{667}.$ (ii) If $m \equiv 1 \pmod{3},$ then by the same reasoning, the smallest positive value of $k$ is $3 - \frac{m}{\lceil \frac{m}{3} \rceil} = 3 - \frac{m}{\frac{m+2}{3}} = 3 - 3 \cdot \frac{m}{m+2}.$ $\frac{m}{m+2}$ increases, approaching $1$, as $m$ increases, so to minimize $3 - 3 \cdot \frac{m}{m+2},$ we want $m$ to be the largest number that is one more than a multiple of $3$ and less than or equal to $2000$ which is $1999.$ If $m=1999,$ this quantity is $\frac{2}{667}.$ (iii) If $m \equiv 2 \pmod{3},$ then by the same reasoning, the smallest positive value of $k$ is $3 - \frac{m}{\lceil \frac{m}{3} \rceil} = 3 - \frac{m}{\frac{m+1}{3}} = 3 - 3 \cdot \frac{m}{m+1}.$ $\frac{m}{m+1}$ increases, approaching $1$, as $m$ increases, so to minimize $3 - 3 \cdot \frac{m}{m+1},$ we want $m$ to be the largest number that is two more than a multiple of $3$ and less than or equal to $2000$ which is $2000.$ If $m=2000,$ this quantity is $\frac{1}{667}.$ Hence, as we have considered all possible cases (the three possible residues modulo $3$), the smallest possible positive value of $k$ is $\boxed{ \frac{1}{667}} .$ $\square$