Why is there negative sign as well as positive sign with currents flowing into node for KCL
Signs of current in a circuit are just conventions chosen at your convenience.
I can say a current of \$1A\$ flows from node A to B. I can also say that current of \$-1A\$ flows from node B to A. Both assumptions are correct. Magnitude and sign, both matter.
Consider following circuit-



simulate this circuit &ndash; Schematic created using CircuitLab
In both circuits I assume voltage between A and B, \$ V_{AB}= +V_1\$ which is equal to \$+10V\$
In the first circuit, I assumed: The direction of \$i_1\$ is &quot;downwards&quot; ie., from A to B.
Since \$i_1\$ enters node A and leaves node B, by ohm's law, $$i1 = V_{AB}/R_1= +V_1/R_1=+10/2=+5A$$
In the second circuit, I assumed: The direction of \$i_1\$ is &quot;upwards&quot; ie., from B to A.
Since \$i_1\$ enters node B and leaves node A, by ohm's law, $$i1 = V_{BA}/R_1= -V_1/R_1=-10/2=-5A$$
So which assumption and calculation was &quot;correct&quot;? Both are correct. Just sign of current changes if you assume its direction to be opposite.
Apply this concept to your circuit in the question-.
Author assumed the directions of currents at his convenience in the circuit. Then,
$$i_1=-v_1/25$$
$$i_4=-v_1/100$$
$$i_{10}=-v1/10$$
Author also assumes that all currents entering node B as positive, and all currents leaving node B as negative. Again, this is his choice, not a rule.
With this assumptions, apply KCL at node B-
$$i_2+i_4+-i_{B\to D}=0$$
$$(i_1+i_{10}+i_{C\to A})+i_4+-i_{B\to D}=0$$
$$\implies \frac{-v_1}{25}+\frac{-v_1}{10}+0.2v_1+\frac{-v_1}{100}+-2.5=0$$