Call $X$ the set containing the $1999$ people and label its elements as $a_{1}, a_{2}, \cdots, a_{1999}$. Now choose one of them, wlog $a_{1}$. By the hypotesis we know that for any $50$ elements there are two people that don't know each other. So there are at least $\lfloor\frac{1998}{49}\rfloor=40$ people that $a_{1}$ doesn't know, or, equivalently, $1958$ people at most that he knows. So let's call $Y$ the subset of $X$ containing $a_{1}$ all the people that $a_{1}$ doesn't know. We immediately have $|Y|=41$. Let's now put in the worst case in which every element $\in Y$ knows any element $\not \in Y$ and there are not three elements of $Y$ in every $50$ people chosen. Thus we should have $\binom{41}{2}$ different couples of elements chosen in $Y$. But it is clear that it is possible to choose a different group of $50$ people for every couple since $\binom{41}{2}\leq \binom{1958}{48}$. To conclude the proof let's dispose the $1999$ people on a circumference and label every $a_{1+k*49}^{th}$ person, with $0\leq k\leq 40$. Every group of $50$ must be chosen in this way. Let's choose $a_{1+k*49}$ and $a_{1+j*49}$ and the $48$ elements preceeding $a_{1+j*49}$. So every group is different from the others and the proof is complete.

venatrix wrote: By the hypotesis we know that for any $50$ elements there are two people that don't know each other. So there are at least $\lfloor\frac{1998}{49}\rfloor=40$ people that $a_{1}$ doesn't know I do not think this follows.

Here is less constructive proof. Observe graph $G$ with $1999$ vertices. Say $X = \{x\in G;d(x)\leq 1958\}$ and $Y = \{x\in G;d(x)\geq 1959\}$. We have to prove $|X| \geq 41$. Assume $|X|\leq 40$, then $|Y|\geq 1959$. Since there is at most $|X|\cdot 1958 \leq 40\cdot 1958$ edges between $X$ and $Y$ there is, by Handshake lemma, at least \[ {1\over 2}(|Y|\cdot 1959 - |X|\cdot 1958) \geq {1\over 2}(1959^2- 40\cdot 1958)\] edges between vertices in $Y$. But since there is no $K_{50}$ in $G$ and there for no $K_{50}$ in $Y$ we have, by Turan theorem, at most: \[ {48\over 2\cdot 49}\cdot 1959^2\] edges in $Y$. So we have (after some transformations) $1959^2\leq 1960\cdot 1958 = 1959^2-1$ which is impossible. So $|X| \geq 41$.

Suppose otherwise. Then we have a minimum of $1959$ people who do not know at most $39$ people. Let $S$ be a set of $1959$ people all who do not know at most $39$ people. Let $Q_i$ be the set which contains all the people that a person $P_i$ does not know.$|Q_i|\le 39$ Let $G_n$ be a group of people such that if $P_i\in G_n$ and $P_j\in G_n$, then $P_i$ knows $P_j$. Clearly $G_2$ exists. If $G_n$ exists with people $P_1,P_2,\ldots,P_n$, we consider the set $T=S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)$. If $T$ has at least one person $P_{n+1}$, we include it in $G$ to give us $G_{n+1}$. But $|T|=|S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)|\ge |S|-|Q_1|-|Q_2|-\cdots-|Q_n|\ge 1959-39n\ge 1$ for $2\ge n\ge 49$. So we have a valid $G_{50}$. Contradiction. EDIT: thanks to some external guidance, here is the more accurate version: If $G_n\subset S$ exists with people $P_1,P_2,\ldots,P_n$, we consider the set $T=S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n\cup G_n)$. If $T$ has at least one person $P_{n+1}$, we include it in $G$ to give us $G_{n+1}\subset S$. But $|T|=|S/(Q_1\cup Q_2\cup Q_3\ldots\cup Q_n)|\ge |S|-|Q_1|-|Q_2|-\cdots-|Q_n|-|G_n|\ge 1959-39n-n\ge 1$ for $2\le n\le 49$ So we have a valid $G_{49}\subset S$. As for $G_{50}$ we just consider some guy not in $Q_1\cup Q_2\cup Q_3\ldots\cup Q_{49}\cup G_{49}$ and add him in, and since $1999-49 \times 39 - 49 \ge 1$ that guy must exist. We have a valid $G_{50}$, contradiction!