Here is the complete classification.

Let

$$ T_n={(a,b)\in\mathbb Z_{\ge1}:\ a+b\le n+1} $$

(the triangular set of lattice points that must be covered). Call a line non‑sunny if it is parallel to one of the three sides of the triangle, i.e. it is of the form $x=\text{const}$, $y=\text{const}$, or $x+y=\text{const}$.

1) A compulsory family of non‑sunny lines

Claim 1. In any admissible family of $n$ lines covering $T_n$, the line

$$ D_{n+1}:\ x+y=n+1 $$

must be chosen.

Proof. The $n$ points $(1,n), (2,n-1),\dots,(n,1)$ are collinear on $D_{n+1}$. If $D_{n+1}$ were not chosen, no other line could contain two of them (the only line through any two is $D_{n+1}$). Hence covering these $n$ points already needs $n$ distinct lines, leaving no line to cover $(1,1)$—a contradiction. ∎

Remove $D_{n+1}$. The remaining $n-1$ lines still cover precisely the points of

$$ T_{n-1}={(a,b)\in\mathbb Z_{\ge1}:\ a+b\le n}, $$

and the number of sunny lines does not change (since $D_{n+1}$ is non‑sunny).

Applying Claim 1 to this smaller triangle shows $D_n:x+y=n$ must also be present; remove it and continue. By induction we obtain:

Corollary. Every admissible family for $T_n$ must contain all the anti‑diagonals

$$ D_5, D_6,\dots, D_{n+1}. $$

Consequently, after fixing those $n-4$ lines, the remaining points to be covered are exactly the six points of the small triangle

$$ T_3={(a,b)\in\mathbb Z_{\ge1}:a+b\le 4}. $$

Only three lines remain to cover $T_3$, and those three (together with the fixed $D_5,\dots,D_{n+1}$) make up the whole admissible family.

So the problem reduces to: Which numbers of sunny lines are possible when three lines must cover $T_3$?

2) The base case $T_3$

The points of $T_3$ are

$$ (1,1),\ (1,2),(2,1),\ (1,3),(2,2),(3,1). $$

• $k=0$ is possible: take the three anti‑diagonals

  $$ D_2:\ x+y=2,\quad D_3:\ x+y=3,\quad D_4:\ x+y=4. $$

• $k=1$ is possible: take $D_3, D_4$ and the sunny line $x=y$ (which covers $(1,1)$ and $(2,2)$).

• $k=3$ is possible: take the three sunny lines

  $$ \ell_1:\ x=y\quad(\text{through }(1,1),(2,2)),\qquad \ell_2:\ \text{through }(1,3)\text{ and }(2,1)\ (,\text{slope }-2,), $$

  $$ \ell_3:\ \text{through }(1,2)\text{ and }(3,1)\ (,\text{slope }-{\tfrac12},). $$

  These three cover all six points of $T_3$.

• $k=2$ is impossible: suppose two sunny lines and one non‑sunny line cover $T_3$. Consider the three possibilities for the non‑sunny line:

  If it is $D_4$: the remaining points are $(1,1),(1,2),(2,1)$. Any line through two of these is vertical, horizontal, or slope $-1$; hence no sunny line contains two of them, so two sunny lines cannot cover all three points.

  If it is $x=1$: the remaining points are $(2,1),(2,2),(3,1)$. Again, any line through two of them is non‑sunny (vertical $x=2$, horizontal $y=1$, or slope $-1$), so two sunny lines cannot cover all three.

  If it is $y=1$: symmetric to the previous case.

  Thus $k=2$ cannot occur for $T_3$.

Therefore, for $T_3$ the possible numbers of sunny lines are exactly $k\in{0,1,3}$.

3) Lifting back to general $n$

As shown above, for general $n$ every admissible family must contain $D_5,\dots,D_{n+1}$, and the remaining three lines must cover $T_3$. Hence the number $k$ of sunny lines for $T_n$ is exactly the number of sunny lines among those three lines for $T_3$. From the base case we conclude that the only possible values are

$$ \boxed{,k\in{0,1,3},}. $$

Finally, explicit constructions for all $n\ge3$:

• $k=0$: take all $n$ non‑sunny anti‑diagonals $x+y=2,3,\dots,n+1$.
• $k=1$: take $x+y=3,4,\dots,n+1$ together with the sunny line $x=y$.
• $k=3$: take $x+y=5,6,\dots,n+1$ together with the three sunny lines $\ell_1,\ell_2,\ell_3$ listed above.

Each set has exactly $n$ distinct lines, covers all points of $T_n$, and has $k=0,1,3$ sunny lines respectively.

Answer: For every integer $n\ge3$, the admissible values of $k$ are precisely $k=0,1,3$.