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I will, as is common, use $v$ rather than $n$ for $|V|.$ The number of edges (often called blocks) is $b=\frac{vr}k$ and this must be an integer. If you are motivated by finite projective planes and the like then you should add a condition that there is a number $\lambda$ such that every pair of vertices is in $\lambda$ common blocks (I will say why in a bit). This number is then determined by $v,k,r$ as $\lambda=\frac{r(k-1)}{v-1}.$ The structure is then called a $2$-design. It is most common to use $(v,k,\lambda)$ as the parameters. It is known that for each fixed pair $k,\lambda$ there is an $N=N_{k,\lambda}$ so that there is a $(v,k,\lambda)$ $2$-design provided that $r=\frac{\lambda(v-1)}{k-1}$ and $b=\frac{vr}k$ are integers. Maybe not that interesting.

Here is one illustration of why this pairwise condition makes it much more interesting: You mention $(v,r,k)=(q^2+q+1,q+1,q+1)$ designs. These are trivial without the pairwise condition. Just pick your $q$ at will. Take $0,1,2,...,v-1$ and the $v$ blocks $\{j,j+1,\cdots,j+k\}$ considered $\mod v.$ That always works. It works for $r=k$ any time $\gcd(v,k)=1.$

With the added condition that $\lambda$ exists, it must be $\lambda=1$ and then one has a finite projective plane of order $q$. There are projective planes (sometimes many) if $q$ is a prime power. A huge question is if there are ever examples with $q$ not a prime power. It is known that if $q=4m+2$ then it must be a sum of two squares for a projective plane of order $q$ to exist. This rules out $q=6$ which is possible to do by hand. A massive calculation ruled out $q=10.$ The case $q=12$ is open.

For the rest of this answer I will forget about the pairwise condition.

If you allow repeated blocks then any $(v,k,r)$ with $\frac{vr}{k} \in \mathbb{Z}$ is possible (provided $k \leq v$) using a simple cyclic construction as above. So let's demand distinct blocks, which limits it to $b \leq \binom{n}{k}.$

My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $v \gt N$ and $b \in \mathbb{Z}.$ In fact you can combine a $(v_1,k,r)$ and $(v_2,k,r)$ with disjoint vertex sets to get a $(v_1+v+2,k,r).$ This allow $(v,k,r)$ for any $v$ of the form $sv_1+tv_2$ which can include all but finitely many possible values of $v.$

I will, as is common, use $v$ rather than $n$ for $|V|.$ The number of edges (often called blocks) is $b=\frac{vr}k$ and this must be an integer. If you are motivated by finite projective planes and the like then you should add a condition that there is a number $\lambda$ such that every pair of vertices is in $\lambda$ common blocks (I will say why in a bit). This number is then determined by $v,k,r$ as $\lambda=\frac{r(k-1)}{v-1}.$ The structure is then called a $2$-design. It is most common to use $(v,k,\lambda)$ as the parameters. It is known that for each fixed pair $k,\lambda$ there is an $N=N_{k,\lambda}$ so that there is a $(v,k,\lambda)$ $2$-design provided that $r=\frac{\lambda(v-1)}{k-1}$ and $b=\frac{vr}k$ are integers. Maybe not that interesting.

Here is one illustration of why this pairwise condition makes it much more interesting: You mention $(v,r,k)=(q^2+q+1,q+1,q+1)$ designs. These are trivial without the pairwise condition. Just pick your $q$ at will. Take $0,1,2,...,v-1$ and the $v$ blocks $\{j,j+1,\cdots,j+k\}$ considered $\mod v.$ That always works. It works for $r=k$ any time $\gcd(v,k)=1.$

With the added condition that $\lambda$ exists, it must be $\lambda=1$ and then one has a finite projective plane of order $q$. There are projective planes (sometimes many) if $q$ is a prime power. A huge question is if there are ever examples with $q$ not a prime power. It is known that if $q=4m+2$ then it must be a sum of two squares for a projective plane of order $q$ to exist. This rules out $q=6$ which is possible to do by hand. A massive calculation ruled out $q=10.$ The case $q=12$ is open.

For the rest of this answer I will forget about the pairwise condition.

If you allow repeated blocks then any $(v,k,r)$ with $\frac{vr}{k} \in \mathbb{Z}$ is possible (provided $k \leq v$) using a simple cyclic construction as above. So let's demand distinct blocks, which limits it to $b \leq \binom{n}{k}.$ I won't go into any detail, but cyclic constructions can cover many cases and these can be combined to cover more.

My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $v \gt N$ and $b \in \mathbb{Z}.$ In fact this may be much easier than in the $2$-design case: You can combine a $(v_1,k,r)$ and $(v_2,k,r)$ with disjoint vertex sets to get a $(v_1+v_2,k,r).$ This allow $(v,k,r)$ for any $v$ of the form $sv_1+tv_2$ which can include all but finitely many possible values of $v.$ Of course this does not tell us about letting $k$ grow with $v.$

The number of edges (often called blocks) is $b=\frac{nr}k$ and this must be an integer. My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $n \gt N$ and that integrality holds. I say this because if you add the condition that there is a $\lambda$ such that every two points are in $\lambda$ common blocks you have a second integrality condition $\lambda(n-1)=r(k-1).$ The structure is then called a $2$-design and a result as I mentioned is known (for fixed $k, \lambda$.)

It may be even simpler than that. You mention $(n,k,r)=(q^2+q+1,q+1,q+1)$ designs. With the added condition that $\lambda$ exists one has a finite projective plane and a huge question is if there are any examples with $q$ not a prime power. The case $q=6$ is ruled out by a number theoretic condition for $q=4m+2.$ Then $q$ must be a sum of two squares. Ruling out $q=10$ was a massive calculation. The case $q=12$ is unknown. BUT without the pairwise condition just pick your $q$ at will. Take $0,1,2,...,n-1$ and the $n$ blocks $\{j,j+1,\cdots,j+k\}$ considered $\mod n.$ That always works.

If you allowed repeated blocks then that kind of cyclic construction might always work. I’d need to think a bit more about the case of no repeated blocks but maybe the integrality condition along with $k\lt n$ and $b \leq \binom{n}k$ suffices.

I will, as is common, use $v$ rather than $n$ for $|V|.$ The number of edges (often called blocks) is $b=\frac{vr}k$ and this must be an integer. If you are motivated by finite projective planes and the like then you should add a condition that there is a number $\lambda$ such that every pair of vertices is in $\lambda$ common blocks (I will say why in a bit). This number is then determined by $v,k,r$ as $\lambda=\frac{r(k-1)}{v-1}.$ The structure is then called a $2$-design. It is most common to use $(v,k,\lambda)$ as the parameters. It is known that for each fixed pair $k,\lambda$ there is an $N=N_{k,\lambda}$ so that there is a $(v,k,\lambda)$ $2$-design provided that $r=\frac{\lambda(v-1)}{k-1}$ and $b=\frac{vr}k$ are integers. Maybe not that interesting.

Here is one illustration of why this pairwise condition makes it much more interesting: You mention $(v,r,k)=(q^2+q+1,q+1,q+1)$ designs. These are trivial without the pairwise condition. Just pick your $q$ at will. Take $0,1,2,...,v-1$ and the $v$ blocks $\{j,j+1,\cdots,j+k\}$ considered $\mod v.$ That always works. It works for $r=k$ any time $\gcd(v,k)=1.$

With the added condition that $\lambda$ exists, it must be $\lambda=1$ and then one has a finite projective plane of order $q$. There are projective planes (sometimes many) if $q$ is a prime power. A huge question is if there are ever examples with $q$ not a prime power. It is known that if $q=4m+2$ then it must be a sum of two squares for a projective plane of order $q$ to exist. This rules out $q=6$ which is possible to do by hand. A massive calculation ruled out $q=10.$ The case $q=12$ is open.

For the rest of this answer I will forget about the pairwise condition.

If you allow repeated blocks then any $(v,k,r)$ with $\frac{vr}{k} \in \mathbb{Z}$ is possible (provided $k \leq v$) using a simple cyclic construction as above. So let's demand distinct blocks, which limits it to $b \leq \binom{n}{k}.$

My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $v \gt N$ and $b \in \mathbb{Z}.$ In fact you can combine a $(v_1,k,r)$ and $(v_2,k,r)$ with disjoint vertex sets to get a $(v_1+v+2,k,r).$ This allow $(v,k,r)$ for any $v$ of the form $sv_1+tv_2$ which can include all but finitely many possible values of $v.$

The number of edges (often called blocks) is $\frac{nr}k$ and this must be an integer. My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $n \gt N$ and that integrality holds. I say this because if you add the condition that there is a $\lambda$ such that every two points are in $\lambda$ common blocks you have a second integrality condition $\lambda(n-1)=r(k-1).$ The structure is then called a $2$-design and a result as I mentioned is known (for fixed $k, \lambda.)

The number of edges (often called blocks) is $b=\frac{nr}k$ and this must be an integer. My guess is that for each fixed $r,k$ there is an $N_{k,r}$ such that what you want exists provided $n \gt N$ and that integrality holds. I say this because if you add the condition that there is a $\lambda$ such that every two points are in $\lambda$ common blocks you have a second integrality condition $\lambda(n-1)=r(k-1).$ The structure is then called a $2$-design and a result as I mentioned is known (for fixed $k, \lambda$.)

It may be even simpler than that. You mention $(n,k,r)=(q^2+q+1,q+1,q+1)$ designs. With the added condition that $\lambda$ exists one has a finite projective plane and a huge question is if there are any examples with $q$ not a prime power. The case $q=6$ is ruled out by a number theoretic condition for $q=4m+2.$ Then $q$ must be a sum of two squares. Ruling out $q=10$ was a massive calculation. The case $q=12$ is unknown. BUT without the pairwise condition just pick your $q$ at will. Take $0,1,2,...,n-1$ and the $n$ blocks $\{j,j+1,\cdots,j+k\}$ considered $\mod n.$ That always works.

If you allowed repeated blocks then that kind of cyclic construction might always work. I’d need to think a bit more about the case of no repeated blocks but maybe the integrality condition along with $k\lt n$ and $b \leq \binom{n}k$ suffices.