Rather than play 20 Questions, here is a solution based on my interpretation of what you posted.
The core of the circuit is one clock oscillator and three CD4017 Johnson Counters in series. The first counter is clocked by an external signal source, the second one is clocked by the Carry output of the first, and the third is clocked by the second. This gets you the 1001 step sequence described. The Q9 output of the third 4017 clocks a flipflop that drives the "IN" led array and inhibits further counting. An R-C power-on reset initializes the circuit, and a switch across the reset capacitor resets the circuit to start another sequence.
A 4017 output can source very little current, so you will need transistor arrays to drive the LEDs. You have 31 distinct outputs; a group of four ULN2803s has 32 outputs and can handle LED currents up to 300 mA reliably.
Update: Just caught this part:
"When the circuit is powered, initially the 1st RED , BLUE and GREEN LEDs will be ON."
This is not as big a problem as it appears. When a 4017 is reset, the Q0 output goes high. This assures that the circuit always starts with the first red, green, and blue LEDs on. Either the Carry Out or the Q9 signal from the third 4017 can be used to clock the external flipflop that drives the IN sign.
The flipflop and the clock oscillator can be made with a single CD4093 quad Schmitt trigger NAND gate.
Having all colored LEDs off when IN is on will take a bit more, because there is no 4017 state where all 10 outputs are low. My first thought here is one PNP transistor, controlled by the IN flipflop, that sources power to the 30 LEDs. What is the power source for the circuit?
Here is a first pass solution. The clock is set to approx. 5 Hz so you can see the LEDs step along. For each of the three lines of LEDs, only one LED is on at a time. That is why there is only one current limiting resistor for 10 LEDs. The correct value for R8 depends on the current requirement of the "IN" array.