Solution:

The set of positive integers sharing a given triple $\left(\left\lfloor \frac{n}{4} \right\rfloor, \left\lfloor \frac{n}{5} \right\rfloor, \left\lfloor \frac{n}{6} \right\rfloor\right)$ is an intersection of three intervals, hence a block of consecutive integers. So $n$ is uniquely determined exactly when neither $n - 1$ nor $n + 1$ gives the same triple: some floor must drop at $n - 1,$ meaning $4, 5,$ or $6$ divides $n,$ and some floor must jump at $n + 1,$ meaning $4, 5,$ or $6$ divides $n + 1.$

Since $n$ and $n + 1$ cannot both be even, the divisor pairs for $(n, n + 1)$ are $(4, 5),$ $(5, 4),$ $(5, 6),$ and $(6, 5).$ Working modulo $60:$ $4 \mid n,\; 5 \mid n + 1$ gives $n \equiv 4, 24, 44;$ $5 \mid n,\; 4 \mid n + 1$ gives $n \equiv 15, 35, 55;$ $5 \mid n,\; 6 \mid n + 1$ gives $n \equiv 5, 35;$ and $6 \mid n,\; 5 \mid n + 1$ gives $n \equiv 24, 54.$ The union is the $8$ residues $\{4, 5, 15, 24, 35, 44, 54, 55\}$ modulo $60.$

Each residue occurs $10$ times among $1 \le n \le 600,$ so the count is $8 \cdot 10 = 80.$ (Note $n = 600$ fails: $601$ is divisible by none of $4, 5, 6,$ so $601, 602, 603$ share $600$'s triple.)