22. The statement remains valid if 17 is replaced by any divisor $k$ of $1989 = 3^2 \cdot 13 \cdot 17$, $1 < k < 1989$, so let $k$ be one such divisor. The set $\{1, 2, \ldots, 1989\}$ can be partitioned as $\{1, 2, \ldots, 3k\} \cup \bigcup_{j=1}^{L} \{(2j+1)k+1, (2j+1)k+2, \ldots, (2j+1)k+2k\} = X \cup Y_1 \cup \cdots \cup Y_L$, where $L = (1989 - 3k) / 2k$. The required statement will be an obvious consequence of the following two claims.

Claim 1. The set $X = \{1, 2, \ldots, 3k\}$ can be partitioned into $k$ disjoint subsets, each having 3 elements and the same sum.

Proof. Since $k$ is odd, let $t = (k-1) / 2$ and $X = \{1, 2, \ldots, 6t+3\}$. For $l = 1, 2, \ldots, t$, define
$$\begin{aligned} X_{2l-1} & = \{l, 3t+1+l, 6t+5-2l\} \\ X_{2l} & = \{t+1+l, 2t+1+l, 6t+4-2l\} \\ X_{2t+1} & = X_k = \{t+1, 4t+2, 4t+3\} \end{aligned}$$
It is easily seen that these three subsets are disjoint and that the sum of elements in each set is $9t+6$.

Claim 2. Each $Y_j = \{(2j+1)k+1, \ldots, (2j+1)k+2k\}$ can be partitioned into $k$ disjoint subsets, each having 2 elements and the same sum.

Proof. The obvious partitioning works:
$$Y_j = \{(2j+1)k+1, (2j+1)k+2k\} \cup \cdots \cup \{(2j+1)k+k, (2j+1)k+(k+1)\}.$$