This line signifies that you are looking at an alignment of your sequence to chain A of structure 1ej0

( https://www.rcsb.org/structure/1EJ0 , FTSJ RNA METHYLTRANSFERASE COMPLEXED WITH S-ADENOSYLMETHIONINE, MERCURY DERIVATIVE).

The aligned part (black) of the sequence corresponds to residues 40 to 209 of the template structure. After modelling, the residues in your model structure will be numbered sequentially, independent of the residue numbering of the template structure. The template structure cannot help in the modelling of the residues indicated in red in your query sequence, and the residues indicated in red in the template structure cannot be aligned to serve as template for these residues.

Dear Annemarie Honegger, How do I find this number of aligned sequences of residues? I did get an alignment file from the previous step using genthreader, but that does not mention any numbers.

If you download and open up file 1ej0.pdb in a text editor, or on https://www.rcsb.org/structure/1EJ0 choose Display Files/PDB format, you can see the content of the pdb file: Each line starts with a keyword denoting the type of information the line contains.

Scroll down to the first line starting with "ATOM":

ATOM 1 N GLY A 30 -10.163 52.668 13.742 1.00 27.29 N

ATOM 2 CA GLY A 30 -8.882 53.486 13.745 1.00 25.22 C

ATOM 3 C GLY A 30 -8.189 53.376 12.395 1.00 20.38 C

ATOM 4 O GLY A 30 -8.535 52.485 11.600 1.00 22.38 O

ATOM 5 N LEU A 31 -7.185 54.218 12.133 1.00 19.22 N

ATOM 6 CA LEU A 31 -6.547 54.133 10.814 1.00 16.25 C

ATOM 7 C LEU A 31 -5.662 52.914 10.663 1.00 14.98 C

ATOM 8 O LEU A 31 -5.461 52.447 9.526 1.00 14.94 O

ATOM 9 CB LEU A 31 -5.727 55.382 10.489 1.00 16.14 C

ATOM 10 CG LEU A 31 -6.367 56.766 10.340 1.00 13.71 C

ATOM 11 CD1 LEU A 31 -5.264 57.778 9.972 1.00 17.57 C

ATOM 12 CD2 LEU A 31 -7.491 56.726 9.284 1.00 20.69 C

ATOM 13 N ARG A 32 -5.117 52.421 11.786 1.00 13.87 N

ATOM 14 CA ARG A 32 -4.251 51.242 11.782 1.00 12.30 C

ATOM 15 C ARG A 32 -4.645 50.342 12.935 1.00 12.83 C

ATOM 16 O ARG A 32 -5.232 50.780 13.924 1.00 14.45 O

ATOM 17 CB ARG A 32 -2.774 51.637 11.958 1.00 12.36 C

ATOM 18 CG ARG A 32 -2.266 52.582 10.870 1.00 9.87 C

ATOM 19 CD ARG A 32 -2.116 51.931 9.502 1.00 10.49 C

ATOM 20 NE ARG A 32 -1.479 52.855 8.565 1.00 12.57 N

ATOM 21 CZ ARG A 32 -2.114 53.766 7.832 1.00 10.80 C

ATOM 22 NH1 ARG A 32 -3.430 53.883 7.888 1.00 10.58 N

ATOM 23 NH2 ARG A 32 -1.408 54.624 7.088 1.00 11.07 N

You can see that the first residue of the structure is Gly A 30, containing atoms N, CA, C and O (four lines). The next residue is Leu A 31, followed by Arg 32 and so on.

The residue numbers do not start with 1, indicating that not all residues in the natural sequence are visible in the structure. If you read the associated publication,

https://www.cell.com/action/showPdf?pii=S1097-2765%2800%2900035-6

you will learn that "Residues 1–29 were not observed in the electron density maps, and amino-terminal sequencing of the protein confirmed that these residues had been cleaved off during the crystallization."