Let the people be dots and the greetings be lines between those dots on a plane. It's easy to see that each greeting counts for two in total, one for one person and one for the other. Assuming n is even we can notice that it's possible for each one to greet exactly 3 other people because if n is congruent to 0 (mod4) we can make sub-groups of 4 in which we know it's possible for each to greet 3 others, and if n is congruent to 2 (mod4) then we can form one sub-group of 6 and then the rest of 4, both of which are possible to configure as the question asks us. Now lets notice that 3n/2 is the amount of greeting there must be since there must be 3 per person but each is shared between 2, and if n were odd then this wouldn't be an interger, which can't be. Therefore it's only possible if n is even. (sorry for not using the LaTeX code notation, AOPS still doesn't let me use it since I'm new)

Consider a graph where each vertex is a person and each edge is a greet. Let $V_1, V_2, ... , V_n$ the vertices. It's well known that $|V_1| + |V_2| + ... + |V_n| = 2A$ where '$A$' is the number of edges (because each edge contains two vertices) and $|V_i|$ is $V_i$'s degree. But we know that $|V_1| + |V_2| + ... + |V_n| = 3n$ because each person to have greeted exactly three other people, so $3n=2A$ and that implies that $2 \mid n$. If $n$ is even: -If $n=4$, connect each vertex with all the 3 vertices. -If $n=6$ connect $V_1 V_2, V_1 V_3, V_1 V_4, V_5 V_2, V_5 V_3, V_5 V_4, V_6 V_2, V_6 V_3, V_6 V_4$ -If $n=4k, k>1$, connect $V_i V_{i+1}, V_i V_{i+2}, V_i V_{i+3}, V_{i+1} V_{i+2}, V_{i+1} V_{i+3}, V_{i+2} V_{i+3}$ for all $i=1, 5, 9, ..., 4t+1, ..., 4k-3$ -If $n=4k+2, k>1$, connect $V_1 V_2, V_1 V_3, V_1 V_4, V_5 V_2, V_5 V_3, V_5 V_4, V_6 V_2, V_6 V_3, V_6 V_4$ and $V_i V_{i+1}, V_i V_{i+2}, V_i V_{i+3}, V_{i+1} V_{i+2}, V_{i+1} V_{i+3}, V_{i+2} V_{i+3}$ for all $i=7, 11, 15, ..., 4t-1, ..., 4k-1$ - - Consideremos un grafo donde cada vértice sea una persona y una arista represente un saludo. Sean $V_1, V_2, ... , V_n$ los vértices. Es bien conocido que: $|V_1| + |V_2| + ... + |V_n| = 2A$ donde $A$ es la cantidad de aristas y $|V_i|$ es el grado de $V_i$. (Debido a que puedes relacionar el mismo saludo dos veces con las manos de dos diferentes personas) Pero, por hipótesis, considerando que cada persona ha saludado a exactamente 3 personas: $|V_1| + |V_2| + ... + |V_n| = 3n$ , por tanto, $3n=2A$ y eso implica que $2 \mid n$. Si $n$ es par: -Si $n=4$, conecta cada vértice con los otros tres vértices. -Si $n=6$ conecta $V_1 V_2, V_1 V_3, V_1 V_4, V_5 V_2, V_5 V_3, V_5 V_4, V_6 V_2, V_6 V_3, V_6 V_4$ -Si $n=4k, k>1$, conecta $V_i V_{i+1}, V_i V_{i+2}, V_i V_{i+3}, V_{i+1} V_{i+2}, V_{i+1} V_{i+3}, V_{i+2} V_{i+3}$ para todo $i=1, 5, 9, ..., 4t+1, ..., 4k-3$ -Si $n=4k+2, k>1$, conecta $V_1 V_2, V_1 V_3, V_1 V_4, V_5 V_2, V_5 V_3, V_5 V_4, V_6 V_2, V_6 V_3, V_6 V_4$ y $V_i V_{i+1}, V_i V_{i+2}, V_i V_{i+3}, V_{i+1} V_{i+2}, V_{i+1} V_{i+3}, V_{i+2} V_{i+3}$ para todo $i=7, 11, 15, ..., 4t-1, ..., 4k-1$

Consider a graph where each person is a node and there's an edge between two nodes iff the corresponding people have greeted each other. If each person has greeted exactly $3$ other people then each vertex has degree $3$, so the sum of the degrees of the vertices is $3n$. Since this must be even (because it's twice the number of edges of the graph), we get that $n$ is even. If $n$ is even, consider a regular $n$-gon and let each vertex greet its two neighbours and the diametrically opposite vertex. In other words, if $n=2m$ and we number the people from $0$ to $2m-1$ then we let the person with number $k$ greet the people with numbers $k-1$, $k+1$ and $k+m$ (indices modulo $2m$).